567 Nice And Hard Inequality - 567 bat dang thuc
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◆❣✉②❡♥ ❉✉② ❚✉♥❣
✺✻✼ ◆✐❝❡ ❆♥❞ ❍❛r❞ ■♥❡q✉❛❧✐t✐❡s
❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂
❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂❂
✶
❚❤✐s ♣r♦❞✉❝t ✐s ❝r❡❛t❡❞ ❢♦r ❡❞✉❝❛t✐♦♥❛❧ ♣✉r♣♦s❡✳
P❧❡❛s❡ ❞♦♥✬t ✉s❡ ✉s❡ ✐t ❢♦r ❛♥② ❝♦♠♠❡❝✐❛❧ ♣✉r♣♦s❡
✉♥❧❡ss ②♦✉ ❣♦t t❤❡ r✐❣❤t ♦❢ t❤❡ ❛✉t❤♦r✳ P❧❡❛s❡ ❝♦♥t❛❝t
❊♠❛✐❧✿♥❣✉②❡♥❞✉②t✉♥❣✾✹❅❣♠❛✐❧✳❝♦♠
✷
✶✳
❛✮ ✐❢ a, b, c ❛r❡ ♣♦s✐t✐✈❡ r❡❛❧ ♥✉♠❜❡rs✱ t❤❡♥
a
b
+
b
c
+
c
a
≥
a
2
+ 1
b
2
+ 1
+
b
2
+ 1
c
2
+ 1
+
c
2
+ 1
a
2
+ 1
.
❜✮▲❡t a, b, c, d ❜❡ ♣♦s✐t✐✈❡ r❡❛❧ ♥✉♠❜❡rs✳Pr♦✈❡ t❤❛t
a
2
− bd
b + 2c + d
+
b
2
− ca
c + 2d + a
+
c
2
− db
d + 2a + b
+
d
2
− ac
a + 2b + c
≥ 0.
❙♦❧✉t✐♦♥✿
❛✮❇② ❈❛✉❝❤②✲❙❝❤✇❛r③✬s ✐♥❡q✉❛❧✐t②✱ ❲❡ ❤❛✈❡✿
a
2
+ b
2
(a
2
+ 1) (b
2
+ 1) ≥
a
2
+ b
2
(ab + 1)
= ab
a
2
+ b
2
+ a
2
+ b
2
≥ ab
a
2
+ b
2
+ 2
⇒
a
b
+
b
a
=
a
2
+ b
2
ab
≥
a
2
+ b
2
+ 2
(a
2
+ 1) (b
2
+ 1)
=
a
2
+ 1
b
2
+ 1
+
b
2
+ 1
a
2
+ 1
❇② ❈❤❡❜②s❤❡✈✬s ✐♥❡q✉❛❧✐t②✱ ❲❡ ❤❛✈❡
a
2
b
2
=
a
2
b
2
+ 1
+
a
2
b
2
(b
2
+ 1)
≥
a
2
b
2
+ 1
+
b
2
b
2
(b
2
+ 1)
=
a
2
+ 1
b
2
+ 1
.
❚❤❡r❡❢♦r❡
1 +
a
b
2
= 1 + 2
a
b
+
b
a
+
a
2
b
2
≥ 1 + 2
a
2
+ 1
b
2
+ 1
+
b
2
+ 1
a
2
+ 1
+
a
2
+ 1
b
2
+ 1
=
1 +
a
2
+ 1
b
2
+ 1
2
.
❚❤❡r❡❢♦r❡
a
b
+
b
c
+
c
a
≥
a
2
+ 1
b
2
+ 1
+
b
2
+ 1
c
2
+ 1
+
c
2
+ 1
a
2
+ 1
❛s r❡q✉✐r❡✳
❜✮◆♦t✐❝❡ t❤❛t
2(a
2
− bd)
b + 2c + d
+ b + d =
2a
2
+ b
2
+ d
2
+ 2c(b + d)
b + 2c + d
=
(a − b)
2
+ (a − d)
2
+ 2(a + c)(b + d)
b + 2c + d
(1)
❆♥❞ s✐♠✐❧❛r❧②✱
2(c
2
− db)
d + 2a + b
+ b + d =
(c − d)
2
+ (c − b)
2
+ 2(a + c)(b + d)
d + 2a + b
(2)
❯s✐♥❣ ❈❛✉❝❤②✲❙❝❤✇❛r③✬s ✐♥❡q✉❛❧✐t②✱✇❡ ❣❡t
(a − d)
2
b + 2c + d
+
(c − d)
2
d + 2a + b
≥
[(a − b)
2
+ (c − d)
2
]
(b + 2c + d) + (d + 2a + b)
(3)
✸
(a − d)
2
b + 2c + d
+
(c − b)
2
d + 2a + b
≥
[(a − d)
2
+ (c − b)
2
]
2
(b + 2c + d) + (d + 2a + b)
(4)
2(a + c)(b + d)
b + 2c + d
+
2(a + c)(b + d)
d + 2a + b
≥
8(a + c)(b + d)
(b + 2c + d) + (d + 2a + b)
(5)
❋r♦♠ ✭✶✮✱✭✷✮✱✭✸✮✱✭✹✮ ❛♥❞ ✭✺✮✱ ✇❡ ❣❡t
2(
a
2
− bd
b + 2c + d
+
c
2
− db
d + 2a + b
) + b + d ≥
(a + c − b − d)
2
+ 4(a + c)(b + d)
a + b + c + d
= a + b + c + d.
♦r
a
2
− bd
b + 2c + d
+
c
2
− db
d + 2a + b
≥
a + c − b − d
2
■♥ t❤❡ s❛♠❡ ♠❛♥♥❡r✱✇❡ ❝❛♥ ❛❧s♦ s❤♦✇ t❤❛t
b
2
− ca
c + 2d + a
+
d
2
− ac
a + 2b + c
≥
b + d − a − c
2
❛♥❞ ❜② ❛❞❞✐♥❣ t❤❡s❡ t✇♦ ✐♥❡q✉❛❧✐t✐❡s✱✇❡ ❣❡t t❤❡ ❞❡s✐r❡❞ r❡s✉❧t✳
❊♥q✉❛❧✐t② ❤♦❧❞s ✐❢ ❛♥❞ ♦♥❧② ✐❢ a = c ❛♥❞ b = d✳
✷✱
▲❡t a, b, c ❜❡ ♣♦s✐t✐✈❡ r❡❛❧ ♥✉♠❜❡rs s✉❝❤ t❤❛t
a + b + c = 1
Pr♦✈❡ t❤❛t t❤❡ ❢♦❧❧♦✇✐♥❣ ✐♥❡q✉❛❧✐t② ❤♦❧❞s
ab
1 − c
2
+
bc
1 − a
2
+
ca
1 − b
2
≤
3
8
❙♦❧✉t✐♦♥✿ ❋r♦♠ t❤❡ ❣✐✈❡♥ ❝♦♥❞✐t✐♦♥ ❚❤❡ ✐♥❡q✉❛❧✐t② ✐s ❡q✉✐✈❛❧❡♥t t♦
4ab
a
2
+ b
2
+ 2(ab + bc + ca)
≤
3
2
❜✉t ❢r♦♠ ❈❛✉❤② ❙❤✇❛r③ ✐♥❡q✉❛❧✐t②
4ab
a
2
+ b
2
+ 2(ab + bc + ca)
≤
ab
a
2
+ ab + bc + ca
+
ab
b
2
+ ab + bc + ca
=
ab
(a + b)(a + c)
+
ab
(b + c)(a + b)
=
a(b + c)
2
(a + b)(b + c)(c + a)
❚❤✉s ❲❡ ♥❡❡❞ ♣r♦✈❡ t❤❛t
3(a + b)(b + c)(c + a) ≥ 2
a(b + c)
2
✇❤✐❝❤ r❡❞✉❝❡s t♦ t❤❡ ♦❜✈✐♦✉s ✐♥❡q✉❛❧✐t②
ab(a + b) ≥ 6abc
❚❤❡ ❙♦❧✉t✐♦♥ ✐s ❝♦♠♣❧❡t❡❞✳✇✐t❤ ❡q✉❛❧✐t② ✐❢ ❛♥❞ ♦♥❧② ✐❢
a = b = c =
1
3
✹
❖r ❲❡ ❝❛♥ ✉s❡ t❤❡ ❢❛❝t t❤❛t
4ab
a
2
+ b
2
+ 2(ab + bc + ca)
≤
4ab
(2ab + 2ac) + (2ab + 2bc)
≤
ab
2a(b + c)
+
ab
2b(a + c)
=
1
2
b
b + c
+
a
a + c
=
1
2
b
b + c
+
c
b + c
=
3
2
✸✱ ▲❡t a, b, c ❜❡ t❤❡ ♣♦s✐t✐✈❡ r❡❛❧ ♥✉♠❜❡rs✳ Pr♦✈❡ t❤❛t
1 +
ab
2
+ bc
2
+ ca
2
(ab + bc + ca)(a + b + c)
≥
4.
3
(a
2
+ ab + bc)(b
2
+ bc + ca)(c
2
+ ca + ab)
(a + b + c)
2
❙♦❧✉t✐♦♥✿ ▼✉❧t✐♣❧②✐♥❣ ❜♦t❤ s✐❞❡s ♦❢ t❤❡ ❛❜♦✈❡ ✐♥❡q✉❛❧✐t② ✇✐t❤ (a + b + c)
2
✐t✬s ❡q✉✐✈❛❧❡♥t t♦
♣r♦✈❡ t❤❛t
(a + b + c)
2
+
(a + b + c)(ab
2
+ bc
2
+ ca
2
)
ab + bc + ca
≥ 4.
3
(a
2
+ ab + bc)(b
2
+ bc + ca)(c
2
+ ca + ab)
❲❡ ❤❛✈❡
(a + b + c)
2
+
(a + b + c)(ab
2
+ bc
2
+ ca
2
)
ab + bc + ca
=
(a
2
+ ab + bc)(c + a)(c + b)
ab + bc + ca
❇② ✉s✐♥❣ ❆▼✲●▼ ✐♥❡q✉❛❧✐t② ❲❡ ❣❡t
(a
2
+ ab + bc)(c + a)(c + b)
ab + bc + ca
≥ 3.
3
(a
2
+ ab + bc)(b
2
+ bc + ca)(c
2
+ ca + ab)[(a + b)(b + c)(c + a)]
2
ab + bc + ca
❙✐♥❝❡ ✐t✬s s✉❢❢✐❝❡s t♦ s❤♦✇ t❤❛t
√
3.
3
(a + b)(b + c)(c + a) ≥ 2.
√
ab + bc + ca
✇❤✐❝❤ ✐s ❝❧❡❛r❧② tr✉❡ ❜② ❆▼✲●▼ ✐♥❡q✉❛❧✐t② ❛❣❛✐♥✳ ❚❤❡ ❙♦❧✉t✐♦♥ ✐s ❝♦♠♣❧❡t❡❞✳ ❊q✉❛❧✐t②
❤♦❧❞s ❢♦r a = b = c
✹✱
▲❡t a
0
, a
1
, . . . , a
n
❜❡ ♣♦s✐t✐✈❡ r❡❛❧ ♥✉♠❜❡rs s✉❝❤ t❤❛t a
k+1
−a
k
≥ 1 ❢♦r ❛❧❧ k = 0, 1, . . . , n−1.
Pr♦✈❡ t❤❛t
1 +
1
a
0
1 +
1
a
1
− a
0
···
1 +
1
a
n
− a
0
≤
1 +
1
a
0
1 +
1
a
1
···
1 +
1
a
n
❙♦❧✉t✐♦♥✿ ❲❡ ✇✐❧❧ ♣r♦✈❡ ✐t ❜② ✐♥❞✉❝t✐♦♥✳
❋♦r n = 1 ❲❡ ♥❡❡❞ t♦ ❝❤❡❝❦ t❤❛t
1 +
1
a
0
1 +
1
a
1
− a
0
≤
1 +
1
a
0
1 +
1
a
1
✇❤✐❝❤ ✐s ❡q✉✐✈❛❧❡♥t t♦ a
0
(a
1
− a
0
− 1) ≥ 0, ✇❤✐❝❤ ✐s tr✉❡ ❜② ❣✐✈❡♥ ❝♦♥❞✐t✐♦♥✳
▲❡t
1 +
1
a
0
1 +
1
a
1
− a
0
···
1 +
1
a
k
− a
0
≤
1 +
1
a
0
1 +
1
a
1
···
1 +
1
a
k
✺
✐t r❡♠❛✐♥s t♦ ♣r♦✈❡ t❤❛t✿
1 +
1
a
0
1 +
1
a
1
− a
0
···
1 +
1
a
k+1
− a
0
≤
≤
1 +
1
a
0
1 +
1
a
1
···
1 +
1
a
k+1
❇② ♦✉r ❤②♣♦t❤❡s✐s
1 +
1
a
0
1 +
1
a
1
···
1 +
1
a
k+1
≥
≥
1 +
1
a
k+1
1 +
1
a
0
1 +
1
a
1
− a
0
···
1 +
1
a
k
− a
0
✐❞ ❡st✱ ✐t r❡♠❛✐♥s t♦ ♣r♦✈❡ t❤❛t✿
1 +
1
a
k+1
1 +
1
a
0
1 +
1
a
1
− a
0
···
1 +
1
a
k
− a
0
≥
≥ 1 +
1
a
0
1 +
1
a
1
− a
0
···
1 +
1
a
k+1
− a
0
❇✉t
1 +
1
a
k+1
1 +
1
a
0
1 +
1
a
1
− a
0
···
1 +
1
a
k
− a
0
≥
≥ 1 +
1
a
0
1 +
1
a
1
− a
0
···
1 +
1
a
k+1
− a
0
⇔
⇔
1
a
k+1
+
1
a
k+1
a
0
1 +
1
a
1
− a
0
···
1 +
1
a
k
− a
0
≥
≥
1
(a
k+1
− a
0
)a
0
1 +
1
a
1
− a
0
···
1 +
1
a
k
− a
0
⇔
⇔ 1 ≥
1
a
k+1
− a
0
1 +
1
a
1
− a
0
···
1 +
1
a
k
− a
0
❇✉t ❜② ♦✉r ❝♦♥❞✐t✐♦♥s ❲❡ ♦❜t❛✐♥✿
1
a
k+1
− a
0
1 +
1
a
1
− a
0
···
1 +
1
a
k
− a
0
≤
≤
1
k
1 +
1
1
···
1 +
1
k − 1
= 1.
❚❤✉s✱ t❤❡ ✐♥❡q✉❛❧✐t② ✐s ♣r♦✈❡♥✳
✺✱
●✐✈❡♥ a, b, c > 0✳ Pr♦✈❡ t❤❛t
3
a
2
+ bc
b
2
+ c
2
≥ 9.
3
√
abc
(a + b + c)
❙♦❧✉t✐♦♥ ✿ ❚❤✐s ✐♥❡q ✐s ❡q✉✐✈❛❧❡♥t t♦✿
a
2
+ bc
3
abc(a
2
+ bc)
2
(b
2
+ c
2
)
≥
9
(a + b + c)
3
❇② ❆▼✲●▼ ✐♥❡q ✱ ❲❡ ❤❛✈❡
a
2
+ bc
3
abc(a
2
+ bc)
2
(b
2
+ c
2
)
=
✻
=
a
2
+ bc
3
(a
2
+ bc)c(a
2
+ bc)b(b
2
+ c
2
)a
≥
3(a
2
+ bc)
sym
a
2
b
❙✐♠✐❧❛r❧②✱ t❤✐s ✐♥❡q ✐s tr✉❡ ✐❢ ❲❡ ♣r♦✈❡ t❤❛t✿
3(a
2
+ b
2
+ c
2
+ ab + bc + ca)
sym
a
2
b
≥
9
(a + b + c)
3
a
3
+ b
3
+ c
3
+ 3abc ≥
sym
a
2
b
❲❤✐❝❤ ✐s tr✉❡ ❜② ❙❝❤✉r ✐♥❡q✳ ❊q✉❛❧✐t② ❤♦❧❞s ✇❤❡♥ a = b = c
✻✱
▲❡t a, b, c ❜❡ ♥♦♥♥❡❣❛t✐✈❡ r❡❛❧ ♥✉♠❜❡rs s✉❝❤ t❤❛t ab + bc + ca > 0✳ Pr♦✈❡ t❤❛t
1
2a
2
+ bc
+
1
2b
2
+ ca
+
1
2c
2
+ ab
≥
2
ab + bc + ca
.
❚❤❡ ✐♥❡q✉❛❧✐t② ✐s ❡q✉✐✈❛❧❡♥t t♦
ab + bc + ca
2a
2
+ bc
≥ 2, (1)
♦r
a(b + c)
2a
2
+ bc
+
bc
bc + 2a
2
≥ 2.(2)
❯s✐♥❣ t❤❡ ❈❛✉❝❤②✲❙❝❤✇❛r③ ✐♥❡q✉❛❧✐t②✱ ❲❡ ❤❛✈❡
bc
bc + 2a
2
≥
(
bc)
2
bc(bc + 2a
2
)
= 1.(3)
❚❤❡r❡❢♦r❡✱ ✐t s✉❢❢✐❝❡s t♦ ♣r♦✈❡ t❤❛t
a(b + c)
2a
2
+ bc
≥ 1.(4)
❙✐♥❝❡
a(b + c)
2a
2
+ bc
≥
a(b + c)
2(a
2
+ bc)
✐t ✐s ❡♥♦✉❣❤ t♦ ❝❤❡❝❦ t❤❛t
a(b + c)
a
2
+ bc
≥ 2, (5)
✇❤✐❝❤ ✐s ❛ ❦♥♦✇♥ r❡s✉❧t✳
❘❡♠❛r❦✿
2ca + bc
2a
2
+ bc
+
2bc + ca
2b
2
+ ca
≥
4c
a + b + c
.
✼✱
▲❡t a, b, c ❜❡ ♥♦♥ ♥❡❣❛t✐✈❡ r❡❛❧ ♥✉♠❜❡rs s✉❝❤ t❤❛t ab + bc + ca > 0✳ Pr♦✈❡ t❤❛t
1
2a
2
+ bc
+
1
2b
2
+ ca
+
1
2c
2
+ ab
+
1
ab + bc + ca
≥
12
(a + b + c)
2
.
❙♦❧✉t✐♦♥✿ ✶✮ ❲❡ ❝❛♥ ♣r♦✈❡ t❤✐s ✐♥❡q✉❛❧✐t② ✉s✐♥❣ t❤❡ ❢♦❧❧♦✇✐♥❣ ❛✉①✐❧✐❛r② r❡s✉❧t
✐❢ 0 ≤ a ≤ min{a, b}✱ t❤❡♥
1
2a
2
+ bc
+
1
2b
2
+ ca
≥
4
(a + b)(a + b + c)
.
✼
✐♥ ❢❛❝t✱ t❤✐s ✐s ✉s❡❞ t♦ r❡♣❧❛❝❡❞ ❢♦r ✧♥♦ t✇♦ ♦❢ ✇❤✐❝❤ ❛r❡ ③❡r♦✧✱ s♦ t❤❛t t❤❡ ❢r❛❝t✐♦♥s
1
2a
2
+ bc
,
1
2b
2
+ ca
,
1
2c
2
+ ab
,
1
ab + bc + ca
❤❛✈❡ ♠❡❛♥✐♥❣s✳
❇❡s✐❞❡s✱ t❤❡ ✐❛❦❡r ❛❧s♦ ✇♦r❦s ❢♦r ✐t✿
1
2a
2
+ bc
+
1
2b
2
+ ca
+
1
2c
2
+ ab
≥
2(ab + bc + ca)
a
2
b
2
+ abc(a + b + c)
❇✉t ♦✉r ❙♦❧✉t✐♦♥ ❢♦r ❜♦t❤ ♦❢ t❤❡♠ ✐s ❡①♣❛♥❞
▲❡t a, b, c ❜❡ ♥♦♥ ♥❡❣❛t✐✈❡ r❡❛❧ ♥✉♠❜❡rs s✉❝❤ t❤❛t ab + bc + ca > 0✳ Pr♦✈❡ t❤❛t
1
2a
2
+ bc
+
1
2b
2
+ ca
+
1
2c
2
+ ab
+
1
ab + bc + ca
≥
12
(a + b + c)
2
.
✷✮ ❈♦♥s✐❞❡r ❜② ❆▼✲●▼ ✐♥❡q✉❛❧✐t②✱ ❲❡ ❤❛✈❡
2
a
2
+ ab + b
2
(a + b + c)
= (2b + a)
2a
2
+ bc
+ (2a + b)
2b
2
+ ca
≥ 2
(2a + b)(2b + a) (2a
2
+ bc) (2b
2
+ ca).
❆♥❞ ❜② ❆▼✲●▼ ✐♥❡q✉❛❧✐t②✱ ❲❡ ❤❛✈❡
c
2
(2a + b)
2a
2
+ bc
+
c
2
(2b + a)
2b
2
+ ca
≥ 2
c
4
(2a + b)(2b + a)
(2a
2
+ bc) (2b
2
+ ca)
≥
2c
2
(2a + b)(2b + a)
(a
2
+ ab + b
2
) (a + b + c)
=
4c
2
a + b + c
+
6abc
a + b + c
c
a
2
+ ab + b
2
2c
2
a + bc
2
+ 2ab
2
+ b
2
c
2a
2
+ bc
=
c
2
(2a + b)
2a
2
+ bc
+
c
2
(2b + a)
2b
2
+ ca
≥
4c
2
a + b + c
+
6abc
a + b + c
c
a
2
+ ab + b
2
=
4
a
2
+ b
2
+ c
2
a + b + c
+
6abc
a + b + c
c
a
2
+ ab + b
2
≥
4
a
2
+ b
2
+ c
2
a + b + c
+
6abc
a + b + c
(a + b + c)
2
c (a
2
+ ab + b
2
)
=
4
a
2
+ b
2
+ c
2
ab + c
+
6abc
ab + bc + ca
⇒
2a
2
b + 2ab
2
+ 2b
2
c + 2bc
2
+ 2c
2
a + 2ca
2
2a
2
+ bc
✽
=
(b + c) +
2c
2
a + bc
2
+ 2ab
2
+ b
2
c
2a
2
+ bc
≥
(b + c) +
4
a
2
+ b
2
+ c
2
a + b + c
+
6abc
ab + bc + ca
=
8
a
2
+ b
2
+ c
2
+ ab + bc + ca
a + b + c
−
2
a
2
b + ab
2
ab + bc + ca
⇒
1
2a
2
+ bc
+
1
ab + bc + ca
≥
4
a
2
+ b
2
+ c
2
+ ab + bc + ca
(a + b + c) (
(a
2
b + ab
2
))
≥
12
(a + b + c)
2
.
<=>
(a + b)(a + c)
2a
2
+ bc
+
a
2
+ bc
2a
2
+ bc
− 2 ≥
12(ab + bc + ca)
(a + b + c)
2
❋r♦♠
2a
2
+ 2bc
2a
2
+ bc
− 3 =
bc
2a
2
+ bc
≥ 1
❲❡ ❣❡t
a
2
+ bc
2a
2
+ bc
− 2 ≥ 0
◆♦✇✱ ❲❡ ✇✐❧❧ ♣r♦✈❡ t❤❡ str♦♥❣❡r
(a + b)(a + c)
2a
2
+ bc
≥
12(ab + bc + ca)
(a + b + c)
2
❋r♦♠ ❝❛✉❝❤②✲s❝❤❛r③t✱ ❲❡ ❤❛✈❡
(a + b)(a + c)
2a
2
+ bc
= (a+b)(b+c)(c+a)(
1
(2a
2
+ bc)(b + c)
≥
3(a + b)(b + c)(c + a)
ab(a + b) + bc(b + c) + ca(c + a)
❋✐♥❛❧❧②✱ ❲❡ ♦♥❧② ♥❡❡❞ t♦ ♣r♦✈❡ t❤❛t
(a + b)(b + c)(c + a)
ab(a + b) + bc(b + c) + ca(c + a)
≥
4(ab + bc + ca)
(a + b + c)
2
(a + b + c)
2
ab + bc + ca
≥
4[ab(a + b) + bc(b + c) + ca(c + a)
(a + b)(b + c)(c + a)
= 4 −
8abc
(a + b)(b + c)(c + a)
a
2
+ b
2
+ c
2
ab + bc + ca
+
8abc
(a + b)(b + c)(c + a)
≥ 2
✇❤✐❝❤ ✐s ♦❧❞ ♣r♦❜❧❡♠✳ ❖✉r ❙♦❧✉t✐♦♥ ❛r❡ ❝♦♠♣❧❡t❡❞ ❡q✉❛❧✐t② ♦❝❝✉r ✐❢ ❛♥❞ ✐❢ ♦♥❧②
a = b = c, a = b, c = 0
♦r ❛♥② ❝②❝❧✐❝ ♣❡r♠✉t✐♦♥✳
✽✱ ▲❡t a, b, c ❜❡ ♣♦s✐t✐✈❡ r❡❛❧ ♥✉♠❜❡rs s✉❝❤ t❤❛t 16(a + b + c) ≥
1
a
+
1
b
+
1
c
✳ Pr♦✈❡ t❤❛t
1
a + b +
2(a + c)
3
≤
8
9
.
❙♦❧✉t✐♦♥✿ ❚❤✐s ♣r♦❜❧❡♠ ✐s r❛t❤❡r ❡❛s②✳ ❯s✐♥❣ t❤❡ ❆▼✲●▼ ✐♥❡q✉❛❧✐t②✱ ❲❡ ❤❛✈❡✿
a + b +
2(c + a) = a + b +
c + a
2
+
c + a
2
≥ 3
3
(a + b)(c + a)
2
.
✾
❙♦ t❤❛t✿
1
a + b +
2(c + a)
3
≤
2
27(a + b)(c + a)
.
❚❤✉s✱ ✐t✬s ❡♥♦✉❣❤ t♦ ❝❤❡❝❦ t❤❛t✿
1
3(a + b)(c + a)
≤ 4 ⇐⇒ 6(a + b)(b + c)(c + a) ≥ a + b + c,
✇❤✐❝❤ ✐s tr✉❡ s✐♥❝❡
9(a + b)(b + c)(c + a) ≥ 8(a + b + c)(ab + bc + ca)
❛♥❞
16abc(a + b + c) ≥ ab + bc + ca ⇒
16(ab + bc + ca)
2
3
≥ ab + bc + ca ⇐⇒ ab + bc + ca ≥
3
16
.
❚❤❡ ❙♦❧✉t✐♦♥ ✐s ❝♦♠♣❧❡t❡❞✳ ❊q✉❛❧✐t② ❤♦❧❞s ✐❢ ❛♥❞ ♦♥❧② ✐❢ a = b = c =
1
4
✳
✾✱ ▲❡t x, y, z ❜❡ ♣♦s✐t✐✈❡ r❡❛❧ ♥✉♠❜❡rs s✉❝❤ t❤❛t xyz = 1✳ Pr♦✈❡ t❤❛t
x
3
+ 1
x
4
+ y + z
+
y
3
+ 1
y
4
+ z + x
+
z
3
+ 1
z
4
+ x + y
≥ 2
√
xy + yz + zx.
❙♦❧✉t✐♦♥✿ ❯s✐♥❣ t❤❡ ❆▼✲●▼ ✐♥❡q✉❛❧✐t②✱ ❲❡ ❤❛✈❡
2
(x
4
+ y + z)(xy + yz + zx) = 2
[x
4
+ xyz(y + z)](xy + yz + zx)
= 2
(x
3
+ y
2
z + yz
2
)(x
2
y + x
2
z + xyz)
≤ (x
3
+ y
2
z + yz
2
) + (x
2
y + x
2
z + xyz)
= (x + y + z)(x
2
+ yz) =
(x + y + z)(x
3
+ 1)
x
.
✐t ❢♦❧❧♦✇s t❤❛t
x
3
+ 1
x
4
+ y + z
≥
2x
√
xy + yz + zx
x + y + z
.
❆❞❞✐♥❣ t❤✐s ❛♥❞ ✐t ❛♥❛❧♦❣♦✉s ✐♥❡q✉❛❧✐t✐❡s✱ t❤❡ r❡s✉❧t ❢♦❧❧♦✇s✳
✶✵✱ ▲❡t a, b, c ❜❡ ♥♦♥♥❡❣❛t✐✈❡ r❡❛❧ ♥✉♠❜❡rs s❛t✐s❢②✐♥❣ a + b + c =
√
5✳ Pr♦✈❡ t❤❛t
(a
2
− b
2
)(b
2
− c
2
)(c
2
− a
2
) ≤
√
5
❙♦❧✉t✐♦♥✿ ❋♦r t❤✐s ♦♥❡✱ ❲❡ ❝❛♥ ❛ss✉♠❡ ❲▲❖● t❤❛t c ≥ b ≥ a s♦ t❤❛t ❲❡ ❤❛✈❡
P = (a
2
− b
2
)(b
2
− c
2
)(c
2
− a
2
) = (c
2
− b
2
)(c
2
− a
2
)(b
2
− a
2
) ≤ b
2
c
2
(c
2
− b
2
).
❆❧s♦ ♥♦t❡ t❤❛t
√
5 = a + b + c ≥ b + c s✐♥❝❡ a ≥ 0✳ ◆♦✇✱ ✉s✐♥❣ t❤❡ ❆▼✲●▼ ✐♥❡q✉❛❧✐t② ❲❡
❤❛✈❡
(c + b) ·
√
5
2
− 1
· c
2
·
√
5
2
+ 1
b
2
· (c − b)
≤ (c + b)
√
5(b + c)
5
5
≤
√
5;
❙♦ t❤❛t ❲❡ ❣❡t P ≤
√
5✳ ❆♥❞ ❤❡♥❝❡ ❲❡ ❛r❡ ❞♦♥❡✳ ❊q✉❛❧✐t② ❤♦❧❞s ✐❢ ❛♥❞ ♦♥❧② ✐❢ (a, b, c) =
√
5
2
+ 1;
√
5
2
− 1; 0
❛♥❞ ❛❧❧ ✐ts ❝②❝❧✐❝ ♣❡r♠✉t❛t✐♦♥s✳ ✷
✶✵
✶✶✱ ▲❡t a, b, c > 0 ❛♥❞ a + b + c = 3✳ Pr♦✈❡ t❤❛t
1
3 + a
2
+ b
2
+
1
3 + b
2
+ c
2
+
1
3 + c
2
+ a
2
≤
3
5
❙♦❧✉t✐♦♥✿ ❲❡ ❤❛✈❡✿
1
3 + a
2
+ b
2
+
1
3 + b
2
+ c
2
+
1
3 + c
2
+ a
2
≤
3
5
<=>
3
3 + a
2
+ b
2
+
3
3 + b
2
+ c
2
+
3
3 + c
2
+ a
2
≤
9
5
a
2
+ b
2
3 + a
2
+ b
2
≥
6
5
❯s✐♥❣ ❈❛✉❝❤②✲❙❝❤✇❛r③✬s ✐♥❡q✉❛❧✐t②✿
a
2
+ b
2
3 + a
2
+ b
2
(
3 + a
2
+ b
2
) ≥ (
a
2
+ b
2
)
2
❚❤❛t ♠❡❛♥s ❲❡ ❤❛✈❡ t♦ ♣r♦✈❡
(
a
2
+ b
2
)
2
≥
6
5
(
(3 + a
2
+ b
2
))
(a
2
+ b
2
) + 2
(a
2
+ b
2
)(a
2
+ c
2
) ≥
54
5
+
12
5
a
2
8
a
2
+ 10
ab ≥ 54 <=> 5(a + b + c)
2
+ 3
a
2
≥ 54
✐t ✐s tr✉❡ ✇✐t❤ a + b + c = 3✳
✶✷✱
●✐✈❡♥ a, b, c > 0 s✉❝❤ t❤❛t ab + bc + ca = 1✳ Pr♦✈❡ t❤❛t
1
4a
2
− bc + 1
+
1
4b
2
− ca + 1
+
1
4c
2
− ab + 1
≥ 1
❙♦❧✉t✐♦♥✿ ✐♥ ❢❛❝t✱ t❤❡ s❤❛r♣❡r ✐♥❡q✉❛❧✐t② ❤♦❧❞s
1
4a
2
− bc + 1
+
1
4b
2
− ca + 1
+
1
4c
2
− ab + 1
≥
3
2
.
❚❤❡ ✐♥❡q✉❛❧✐t② ✐s ❡q✉✐✈❛❧❡♥t t♦
1
a(4a + b + c)
+
1
b(4b + c + a)
+
1
c(4c + a + b)
≥
3
2
.
❯s✐♥❣ t❤❡ ❈❛✉❝❤②✲❙❝❤✇❛r③ ✐♥❡q✉❛❧✐t②✱ ❲❡ ❤❛✈❡
1
a(4a + b + c)
4a + b + c
a
≥
1
a
2
=
1
a
2
b
2
c
2
.
❚❤❡r❡❢♦r❡✱ ✐t s✉❢❢✐❝❡s t♦ ♣r♦✈❡ t❤❛t
2
3a
2
b
2
c
2
≥
4a + b + c
a
+
4b + c + a
b
+
4c + a + b
c
.
❙✐♥❝❡
4a + b + c
a
=
3 +
a + b + c
a
= 9 +
(a + b + c)(ab + bc + ca)
abc
= 9 +
a + b + c
abc
,
✶✶
t❤✐s ✐♥❡q✉❛❧✐t② ❝❛♥ ❜❡ ✇r✐tt❡♥ ❛s
9a
2
b
2
c
2
+ abc(a + b + c) ≤
2
3
,
✇❤✐❝❤ ✐s tr✉❡ ❜❡❝❛✉s❡
a
2
b
2
c
2
≤
ab + bc + ca
3
3
=
1
27
,
❛♥❞
abc(a + b + c) ≤
(ab + bc + ca)
2
3
=
1
3
.
✶✸✱ ●✐✈❡♥ a, b, c ≥ 0 s✉❝❤ t❤❛t ab + bc + ca = 1✳ Pr♦✈❡ t❤❛t
1
4a
2
− bc + 2
+
1
4b
2
− ca + 2
+
1
4c
2
− ab + 2
≥ 1
❙♦❧✉t✐♦♥✿ ◆♦t✐❝❡ t❤❛t t❤❡ ❝❛s❡ abc = 0 ✐s tr✐✈✐❛❧ s♦ ❧❡t ✉s ❝♦♥s✐❞❡r ♥♦✇ t❤❛t abc > 0✳ ❯s✐♥❣
t❤❡ ❆▼✲●▼ ✐♥❡q✉❛❧✐t②✱ ❲❡ ❤❛✈❡
4a
2
− bc + 2(ab + bc + ca) = (2a + b)(2a + c) ≤
[c(2a + b) + b(2a + c)]
2
4bc
=
(ab + bc + ca)
2
bc
=
1
bc
.
✐t ❢♦❧❧♦✇s t❤❛t
1
4a
2
− bc + 2
≥ bc.
❆❞❞✐♥❣ t❤✐s ❛♥❞ ✐ts ❛♥❛❧♦❣♦✉s ✐♥❡q✉❛❧✐t✐❡s✱ ❲❡ ❣❡t t❤❡ ❞❡s✐r❡❞ r❡s✉❧t✳
✶✹✱ ●✐✈❡♥ a, b, c ❛r❡ ♣♦s✐t✐✈❡ r❡❛❧ ♥✉♠❜❡rs✳ Pr♦✈❡ t❤❛t
(
1
a
+
1
b
+
1
c
)(
1
1 + a
+
1
1 + b
+
1
1 + c
) ≥
9
1 + abc
.
❙♦❧✉t✐♦♥✿ ❚❤❡ ♦r✐❣✐♥❛❧ ✐♥❡q✉❛❧✐t② ✐s ❡q✉✐✈❛❧❡♥t t♦
abc + 1
a
+
abc + 1
b
+
abc + 1
c
1
a + 1
+
1
b + 1
+
1
c + 1
≥ 9
♦r
cyc
1 + a
2
c
a
1
a + 1
+
1
b + 1
+
1
c + 1
≥ 9
❇② ❈❛✉❝❤② ❙❝❤✇❛r③ ✐♥❡q ❛♥❞ ❆▼✲●▼ ✐♥❡q✱
cyc
1 + a
2
c
a
≥
cyc
c(1 + a)
2
a(1 + c)
≥ 3
3
(1 + a)(1 + b)(1 + c)
❛♥❞
1
a + 1
+
1
b + 1
+
1
c + 1
≥
3
3
(1 + a)(1 + b)(1 + c)
▼✉❧t✐♣❧②✐♥❣ t❤❡s❡ t✇♦ ✐♥❡q✉❛❧✐t✐❡s✱ t❤❡ ❝♦♥❝❧✉s✐♦♥ ❢♦❧❧♦✇s✳ ❊q✉❛❧✐t② ❤♦❧❞s ✐❢ ❛♥❞ ♦♥❧② ✐❢
a = b = c = 1✳
✶✺✳ ●✐✈❡♥ a, b, c ❛r❡ ♣♦s✐t✐✈❡ r❡❛❧ ♥✉♠❜❡rs✳ Pr♦✈❡ t❤❛t✿
a(b + 1) +
b(c + 1) +
c(a + 1) ≤
3
2
(a + 1)(b + 1)(c + 1)
✶✷
❙♦❧✉t✐♦♥✿ ❈❛s❡✶✳✐❢ a + b + c + ab + bc + ca ≤ 3abc + 3 <=> 4(ab + bc + ca + a + b + c) ≤
3(a + 1)(b + 1)(c + 1) ❯s✐♥❣ ❈❛✉❝❤②✲❙❝❤❛✇r③✬s ✐♥❡q✉❛❧✐t② ✱❲❡ ❤❛✈❡✿
(
a(b + 1) +
b(c + 1) +
c(a + 1))
2
≤ 3(ab + bc + ca + a + b + c) ≤
9(a + 1)(b + 1)(c + 1)
4
❚❤❡ ✐♥❡q✉❛❧✐t② ✐s tr✉❡✳ ❈❛s❡✷✳ ✐❢a + b + c + ab + bc + ca ≤ 3abc + 3.
<=>
9(a + 1)(b + 1)(c + 1)
4
≥ 2(a + b + c + ab + bc + ca) + 3abc + 3
❇② ❆▼✲●▼✬s ✐♥❡q✉❛❧✐t② ✿
2
ab(b + 1)(c + 1) ≤
[ab(c + 1) + (b + 1)] = a + b + c + ab + bc + ca + 3abc + 3
=> ab + bc + ca + a + b + c + 2
ab(b + 1)(c + 1) ≤
9
4(a + 1)(b + 1)(c + 1)
=> (
a(b + 1) +
b(c + 1) +
c(a + 1))
2
≤ [
3
2
(a + 1)(b + 1)(c + 1)]
2
=> Q.E.D
❊♥q✉❛❧✐t② ❤♦❧❞s ✇❤❡♥ a = b = c = 1.
✶✻✱ ●✐✈❡♥ a, b, c ❛r❡ ♣♦s✐t✐✈❡ r❡❛❧ ♥✉♠❜❡rs✳ Pr♦✈❡ t❤❛t✿
1
a
2
+ b
2
+
1
b
2
+ c
2
+
1
c
2
+ a
2
≥
10
(a + b + c)
2
❙♦❧✉t✐♦♥✿ ❆ss✉♠❡ c = min{a, b, c}✳ ❚❤❡♥
1
a
2
+ c
2
+
1
b
2
+ c
2
≥
2
ab + c
2
⇐⇒ (ab − c
2
)(a − b)
2
≥ 0
❆♥❞ ❜② ❈❛✉❝❤②✲s❝❤✇❛r③
((a
2
+ b
2
) + 8(ab + c
2
))
1
a
2
+ b
2
+
2
ab + c
2
≥ 25
❍❡♥❝❡ ❲❡ ♥❡❡❞ ♦♥❧② t♦ ♣r♦✈❡✿
5(a + b + c)
2
≥ 2((a
2
+ b
2
) + 8(ab + c
2
)) ⇐⇒
3(a − b)
2
+ c(10b + 10a − 11c) ≥ 0
❊q✉❛❧✐t② ❢♦r a = b, c = 0 ♦r ♣❡r♠✉t❛t✐♦♥s✳
✶✼✱ ▲❡t a, b ❛♥❞ c ❛r❡ ♥♦♥✲♥❡❣❛t✐✈❡ ♥✉♠❜❡rs s✉❝❤ t❤❛t ab + ac + bc = 0✳ Pr♦✈❡ t❤❛t
a
2
(b + c)
2
a
2
+ 3bc
+
b
2
(a + c)
2
b
2
+ 3ac
+
c
2
(a + b)
2
c
2
+ 3ab
≤ a
2
+ b
2
+ c
2
❙♦❧✉t✐♦♥✿
❇② ❈❛✉❝❤②✲❙❝❤✇❛r③ ✐♥❡q ✱ ❲❡ ❤❛✈❡
a
2
(b + c)
2
a
2
+ bc
=
a
2
(b + c)
3
(a
2
+ bc)(b + c)
=
a
2
(b + c)
3
b(a
2
+ c
2
) + c(a
2
+ b
2
)
≤
a
2
(b + c)
4
(
b
2
b(a
2
+ c
2
)
+
c
2
c(a
2
+ b
2
)
) =
a
2
(b + c)
4
(
b
a
2
+ c
2
+
c
a
2
+ b
2
)
❙✐♠✐❧❛r❧②✱ ❲❡ ❤❛✈❡
LHS ≤
a
2
(b + c)(
b
a
2
+ c
2
+
c
a
2
+ b
2
) =
c(a
2
(b + c) + b
2
(c + a))
a
2
+ b
2
✶✸
= a
2
+ b
2
+ c
2
+
abc(a + b)
a
2
+ b
2
≤ a
2
+ b
2
+ c
2
+
abc(a + b)
a
2
+ b
2
≤ a
2
+ b
2
+ c
2
+ ab + bc + ca
✇❤✐❝❤ ✐s tr✉❡ ❜② ❆▼✲●▼ ✐♥❡q
❚❤❡ ♦r✐❣✐♥❛❧ ✐♥❡q✉❛❧✐t② ❝❛♥ ❜❡ ✇r✐tt❡♥ ❛s
(a + b)
2
(a + c)
2
a
2
+ bc
≤
8
3
(a + b + c)
2
.
❙✐♥❝❡ (a + b)(a + c) = (a
2
+ bc) + a(b + c) ❲❡ ❤❛✈❡
(a + b)
2
(a + c)
2
a
2
+ bc
=
(a
2
+ bc)
2
+ 2a(b + c)(a
2
+ bc) + a
2
(b + c)
2
a
2
+ bc
= a
2
+ bc + 2a(b + c) +
a
2
(b + c)
2
a
2
+ bc
,
❛♥❞ t❤✉s t❤❡ ❛❜♦✈❡ ✐♥❡q✉❛❧✐t② ✐s ❡q✉✐✈❛❧❡♥t t♦
a
2
(b + c)
2
a
2
+ bc
≤
8
3
(a + b + c)
2
−
a
2
− 5
ab,
♦r
a
2
(b + c)
2
a
2
+ bc
≤
5(a
2
+ b
2
+ c
2
) + ab + bc + ca
3
.
❙✐♥❝❡
5(a
2
+ b
2
+ c
2
) + ab + bc + ca
3
≥ a
2
+ b
2
+ c
2
+ ab + bc + ca
✐t ✐s ❡♥♦✉❣❤ s❤♦✇ t❤❛t
a
2
(b + c)
2
a
2
+ bc
≤ a
2
+ b
2
+ c
2
+ ab + bc + ca.
◗✳❊✳❉
✶✽✱ ●✐✈❡♥
a
1
≥ a
2
≥ . . . ≥ a
n
≥ 0, b
1
≥ b
2
≥ . . . ≥ b
n
≥ 0
n
i=1
a
i
= 1 =
n
i=1
b
i
❋✐♥❞ t❤❡ ♠❛①♠✐✉♠ ♦❢
n
i=1
(a
i
− b
i
)
2
❲❙♦❧✉t✐♦♥✿✐t❤♦✉t ❧♦ss ♦❢ ❣❡♥❡r❛❧✐t②✱ ❛ss✉♠❡ t❤❛t
a
1
≥ b
1
◆♦t✐❝❡ t❤❛t ❢♦r
a ≥ x ≥ 0, b, y ≥ 0
❲❡ ❤❛✈❡
(a − x)
2
+ (b − y)
2
− (a + b − x)
2
− y
2
= −2b(a − x + y) ≤ 0.
❆❝❝♦r❞✐♥❣ t♦ t❤✐s ✐♥❡q✉❛❧✐t②✱ ❲❡ ❤❛✈❡
(a
1
− b
1
)
2
+ (a
2
− b
2
)
2
≤ (a
1
+ a
2
− b
1
)
2
+ b
2
2
,
(a
1
+ a
2
− b
1
)
2
+ (a
3
− b
3
)
2
≤ (a
1
+ a
2
+ a
3
− b
1
)
2
+ b
2
3
,······
✶✹
(a
1
+ a
2
+ ··· + a
n−1
− b
1
)
2
+ (a
n
− b
n
)
2
≤ (a
1
+ a
2
+ ··· + a
n
− b
1
)
2
+ b
2
n
.
❆❞❞✐♥❣ t❤❡s❡ ✐♥❡q✉❛❧✐t✐❡s✱ ❲❡ ❣❡t
n
i=1
(a
i
− b
i
)
2
≤ (1 − b
1
)
2
+ b
2
2
+ b
2
3
+ ··· + b
2
n
≤ (1 − b
1
)
2
+ b
1
(b
2
+ b
3
+ ··· + b
n
)
= (1 − b
1
)
2
+ b
1
(1 − b
1
) = 1 − b
1
≤ 1 −
1
n
.
❊q✉❛❧✐t② ❤♦❧❞s ❢♦r ❡①❛♠♣❧❡ ✇❤❡♥
a
1
= 1, a
2
= a
3
= ··· = a
n
= 0
❛♥❞
b
1
= b
2
= ··· = b
n
=
1
n
✶✾✱ ●✐✈❡♥
a, b, c ≥ 0
s✉❝❤ t❤❛t
a
2
+ b
2
+ c
2
= 1
Pr♦✈❡ t❤❛t
1 − ab
7 − 3ac
+
1 − bc
7 − 3ba
+
1 − ca
7 − 3cb
≥
1
3
❙♦❧✉t✐♦♥✿ ❋✐rst✱ ❲❡ ✇✐❧❧ s❤♦✇ t❤❛t
1
7 − 3ab
+
1
7 − 3bc
+
1
7 − 3ca
≤
1
2
.
❯s✐♥❣ t❤❡ ❈❛✉❝❤②✲❙❝❤✇❛r③ ✐♥❡q✉❛❧✐t②✱ ❲❡ ❤❛✈❡
1
7 − 3ab
=
1
3(1 − ab) + 4
≤
1
9
1
3(1 − ab)
+ 1
.
✐t ❢♦❧❧♦✇s t❤❛t
1
7 − 3ab
≤
1
27
1
1 − ab
+
1
3
,
❛♥❞ t❤✉s✱ ✐t ✐s ❡♥♦✉❣❤ t♦ s❤♦✇ t❤❛t
1
1 − ab
+
1
1 − bc
+
1
1 − ca
≤
9
2
,
✇❤✐❝❤ ✐s ❱❛s❝✬s ✐♥❡q✉❛❧✐t②✳ ◆♦✇✱ ❲❡ ✇r✐t❡ t❤❡ ♦r✐❣✐♥❛❧ ✐♥❡q✉❛❧✐t② ❛s
3 − 3ab
7 − 3ac
+
3 − 3bc
7 − 3ba
+
3 − 3ca
7 − 3cb
≥ 1,
♦r
7 − 3ab
7 − 3ac
+
7 − 3bc
7 − 3ba
+
7 − 3ca
7 − 3cb
≥ 1 + 4
1
7 − 3ab
+
1
7 − 3bc
+
1
7 − 3ca
.
❙✐♥❝❡
4
1
7 − 3ab
+
1
7 − 3bc
+
1
7 − 3ca
≤ 2
✶✺
✐t ✐s ❡♥♦✉❣❤ t♦ s❤♦✇ t❤❛t
7 − 3ab
7 − 3ac
+
7 − 3bc
7 − 3ba
+
7 − 3ca
7 − 3cb
≥ 3,
✇❤✐❝❤ ✐s tr✉❡ ❛❝❝♦r❞✐♥❣ t♦ t❤❡ ❆▼✲●▼ ✐♥❡q✉❛❧✐t②✳
✷✶✱ ▲❡t
a, b, c ≥ 0
s✉❝❤ t❤❛t
a + b + c > 0
❛♥❞
b + c ≥ 2a
❋♦r
x, y, z > 0
s✉❝❤ t❤❛t
xyz = 1
Pr♦✈❡ t❤❛t t❤❡ ❢♦❧❧♦✇✐♥❣ ✐♥❡q✉❛❧✐t② ❤♦❧❞s
1
a + x
2
(by + cz)
+
1
a + y
2
(bz + cx)
+
1
a + z
2
(bx + cy)
≥
3
a + b + c
❙♦❧✉t✐♦♥✿ ❙❡tt✐♥❣
u =
1
x
, v =
1
y
❛♥❞
w =
1
z
❛♥❞ ✉s✐♥❣ t❤❡ ❝♦♥❞✐t✐♦♥
uvw = 1
t❤❡ ✐♥❡q✉❛❧✐t② ❝❛♥ ❜❡ r❡✇r✐tt❡♥ ❛s
u
au + cv + bw
=
u
2
au
2
+ cuv + bwu
3
a + b + c
✳
❆♣♣❧②✐♥❣ ❈❛✉❝❤②✱ ✐t s✉❢❢✐❝❡s t♦ ♣r♦✈❡
(u + v + w)
2
a
u
2
+ (b + c)
uv
3
a + b + c
1
2
· (b + c − 2a)
(x − y)
2
0✱
✇❤✐❝❤ ✐s ♦❜✈✐♦✉s ❞✉❡ t♦ t❤❡ ❝♦♥❞✐t✐♦♥ ❢♦r
a, b, c
✷✷✱ ●✐✈❡♥
x, y, z > 0
s✉❝❤ t❤❛t
xyz = 1
✶✻
Pr♦✈❡ t❤❛t
1
(1 + x
2
)(1 + x
7
)
+
1
(1 + y
2
)(1 + y
7
)
+
1
(1 + z
2
)(1 + z
7
)
≥
3
4
❙♦❧✉t✐♦♥✿ ❋✐rst ❲❡ ♣r♦✈❡ t❤✐s ✐♥❡q ❡❛s②
1
(1 + x
2
)(1 + x
7
)
≥
3
4(x
9
+ x
9
2
+ 1)
❆♥❞ t❤✐s ✐♥❡q ❜❡❝❛♠❡✿
1
x
9
+ x
9
2
+ 1
+
1
y
9
+ y
9
2
+ 1
+
1
z
9
+ z
9
2
+ 1
≥ 1
✇✐t❤
xyz = 1
✐t✬s ❛♥ ♦❧❞ r❡s✉❧t
✷✸✱ ▲❡t
a, b, c
❜❡ ♣♦s✐t✐✈❡ r❡❛❧ ♥✉♠❜❡rs s✉❝❤ t❤❛t
3(a
2
+ b
2
+ c
2
) + ab + bc + ca = 12
Pr♦✈❡ t❤❛t
a
√
a + b
+
b
√
b + c
+
c
√
c + a
≤
3
√
2
.
❙♦❧✉t✐♦♥✿ ▲❡t
A = a
2
+ b
2
+ c
2
, B = ab + bc + ca
2A + B = 2
a
2
+
ab ≤
3
4
3
a
2
+
ab
= 9.
❇② ❈❛✉❝❤② ❙❝❤✇❛r③ ✐♥❡q✉❛❧✐t②✱ ❲❡ ❤❛✈❡
a
√
a + b
=
√
a
a
a + b
≤
√
a + b + c
a
a + b
.
❇② ❈❛✉❝❤② ❙❝❤✇❛r③ ✐♥❡q✉❛❧✐t② ❛❣❛✐♥✱ ❲❡ ❤❛✈❡
b
a + b
=
b
2
b(a + b)
≥
(a + b + c)
2
b(a + b)
=
A + 2B
A + B
a
a + b
= 3 −
b
a + b
≤ 3 −
A + 2B
A + B
=
2A + B
A + B
❤❡♥❝❡✱ ✐t s✉❢❢✐❝❡s t♦ ♣r♦✈❡ t❤❛t
(a + b + c) ·
2A + B
A + B
≤
9
2
✶✼
❈♦♥s✐❞❡r
(a + b + c)
√
2A + B
=
(A + 2B) (2A + B)
≤
(A + 2B) + (2A + B)
2
=
3
2
(A + B)
⇒ (a + b + c) ·
2A + B
A + B
≤
3
2
√
2A + B ≤
9
2
❛s r❡q✉✐r❡✳
❇② ❆▼✲●▼ ✐♥❡q ❡❛s② t♦ s❡❡ t❤❛t
3 ≤ a
2
+ b
2
+ c
2
≤ 4
❇② ❈❛✉❝❤②✲❙❝❤✇❛r③ ✐♥❡q✱ ❲❡ ❤❛✈❡
LHS
2
= (
a
√
a + c
(a + b)(a + c)
) ≤ (a
2
+ b
2
+ c
2
+ ab + bc + ca)(
a
(a + b)(a + c)
)
❯s✐♥❣ t❤❡ ❢❛♠✐❧✐❛r ✐♥❡q
9(a + b)(b + c)(c + a) ≥ 8(a + b + c)(ab + bc + ca)
❲❡ ❤❛✈❡
a
(a + b)(a + c)
=
2(ab + bc + ca)
(a + b)(b + c)(c + a)
≤
9
4(a + b + c)
❆♥❞ ❲❡ ♥❡❡❞ t♦ ♣r♦✈❡ t❤❛t
9(a
2
+ b
2
+ c
2
+ ab + bc + ca)
4(a + b + c)
≤
9
2
⇔
6 − (a
2
+ b
2
+ c
2
)
24 − 5(a
2
+ b
2
+ c
2
)
≤ 1
⇔ (6 − (a
2
+ b
2
+ c
2
))
2
≤ 24 − 5(a
2
+ b
2
+ c
2
)
⇔ (3 − (a
2
+ b
2
+ c
2
))(4 − (a
2
+ b
2
+ c
2
)) ≤ 0
❲❤✐❝❤ ✐s tr✉❡ ❲❡ ❛r❡ ❞♦♥❡ ❡q✉❛❧✐t② ❤♦❧❞s ✇❤❡♥
a = b = c = 1
✷✹✳
●✐✈❡♥
a, b, c ≥ 0
Pr♦✈❡ t❤❛t
1
(a
2
+ bc)(b + c)
2
≤
8(a + b + c)
2
3(a + b)
2
(b + c)
2
(c + a)
2
❙♦❧✉t✐♦♥✿ ✐♥ ❢❛❝t✱ t❤❡ s❤❛r♣❡r ❛♥❞ ♥✐❝❡r ✐♥❡q✉❛❧✐t② ❤♦❧❞s✿
a
2
(b + c)
2
a
2
+ bc
+
b
2
(c + a)
2
b
2
+ ca
+
c
2
(a + b)
2
c
2
+ ab
≤ a
2
+ b
2
+ c
2
+ ab + bc + ca.
a
2
(b + c)
2
a
2
+ bc
+
b
2
(c + a)
2
b
2
+ ca
+
c
2
(a + b)
2
c
2
+ ab
≤ a
2
+ b
2
+ c
2
+ ab + bc + ca
✶✽
✷✺✳
●✐✈❡♥
a, b, c ≥ 0
s✉❝❤ t❤❛t
ab + bc + ca = 1
Pr♦✈❡ t❤❛t
1
8
5
a
2
+ bc
+
1
8
5
b
2
+ ca
+
1
8
5
c
2
+ ab
≥
9
4
❆ss✉♠❡ ❲▲❖●
a ≥ b ≥ c
t❤✐s ✐♥❡q
1
8
5
a
2
+ bc
−
5
8
+
1
8
5
b
2
+ ca
−
5
8
+
1
8
5
c
2
+ ab
− 1 ≥ 0
8 − 8a
2
− 5bc
8a
2
+ 5bc
+
8 − 8b
2
− 5ca
8b
2
+ 5ca
+
1 −
8
5
c
2
− ab
c
2
+
8
5
ab
≥ 0
8a(b + c − a) + 3bc
8a
2
+ 5bc
+
8b(a + c − b) + 5ac
8b
2
+ 5ca
+
c(a + b −
8
5
c)
c
2
+
8
5
ab
≥ 0
◆♦t✐❝❡ t❤❛t ❲❡ ♦♥❧② ♥❡❡❞ t♦ ♣r♦✈❡ t❤✐s ✐♥❡q ✇❤❡♥
a ≥ b + c
❜② t❤❡ ✇❛② ❲❡ ♥❡❡❞ t♦ ♣r♦✈❡ t❤❛t
8b
8b
2
+ 5ca
≥
8a
8a
2
+ 5bc
(a − b)(8ab − 5ac − 5bc) ≥ 0
❊❛s② t♦ s❡❡ t❤❛t✿ ✐❢
a ≥ b + c
t❤❡♥
8ab = 5ab + 3ab ≥ 5ac + 6bc ≥ 5ac + 5ac
❙♦ t❤✐s ✐♥❡q ✐s tr✉❡✱ ❲❡ ❤❛✈❡ q✳❞✳❡ ✱ ❡q✉❛❧✐t② ❤♦❧❞ ✇❤❡♥
(a, b, c) = (1, 1, 0)
✷✻✱ ●✐✈❡
a, b, c ≥ 0
Pr♦✈❡ t❤❛t✿
a
b
2
+ c
2
+
b
a
2
+ c
2
+
c
a
2
+ b
2
≥
a + b + c
ab + bc + ca
+
abc(a + b + c)
(a
3
+ b
3
+ c
3
)(ab + bc + ca)
a
b
2
+ c
2
=
a
2
ab
2
+ c
2
a
≥
(a + b + c)
2
(ab
2
+ c
2
a)
,
✐t s✉❢❢✐❝❡s t♦ ♣r♦✈❡ t❤❛t
a + b + c
(ab
2
+ c
2
a)
≥
1
ab + bc + ca
+
abc
(ab + bc + ca) (a
3
+ b
3
+ c
3
)
,
✶✾
❜❡❝❛✉s❡
a + b + c
(ab
2
+ c
2
a)
−
1
ab + bc + ca
=
3abc
(ab + bc + ca)
(ab
2
+ ca
2
)
,
✐t s✉❢❢✐❝❡s t♦ ♣r♦✈❡ t❤❛t
3
a
3
+ b
3
+ c
3
≥
ab
2
+ c
2
a
,
✇❤✐❝❤ ✐s tr✉❡ ❜❡❝❛✉s❡
2
a
3
+ b
3
+ c
3
≥
ab
2
+ c
2
a
.
❘❡♠❛r❦✿
a
b
2
+ c
2
+
b
c
2
+ a
2
+
c
a
2
+ b
2
≥
a + b + c
ab + bc + ca
+
3abc(a + b + c)
2(a
3
+ b
3
+ c
3
)(ab + bc + ca)
.
●✐✈❡
a, b, c ≥ 0
Pr♦✈❡ t❤❛t
1
a
2
+ bc
+
1
b
2
+ ca
+
1
c
2
+ ab
≥
3
ab + bc + ca
+
81a
2
b
2
c
2
2(a
2
+ b
2
+ c
2
)
4
❊q✉❛❧✐t② ♦❝❝✉r ✐❢ ❛♥❞ ✐❢ ♦♥❧②
a = b = c, a = b, c = 0
♦r ❛♥② ❝②❝❧✐❝ ♣❡r♠✉t✐♦♥✳
✐t ✐s tr✉❡ ❜❡❝❛✉s❡
(1)
1
a
2
+ bc
+
1
b
2
+ ca
+
1
c
2
+ ab
≥
3
a
2
+ b
2
+ c
2
a
3
b + ab
3
+ b
3
c + bc
3
+ c
3
a + ca
3
❛♥❞
(2)
3
a
2
+ b
2
+ c
2
a
3
b + ab
3
+ b
3
c + bc
3
+ c
3
a + ca
3
≥
3
ab + bc + ca
+
81a
2
b
2
c
2
2(a
2
+ b
2
+ c
2
)
4
.
❇❡❝❛✉s❡
a
2
(a
3
b + ab
3
)
−
1
ab + bc + ca
=
abc(a + b + c)
(ab + bc + ca) (
(a
3
b + ab
3
))
,
✐t s✉❢❢✐❝❡s t♦ ♣r♦✈❡ t❤❛t
2(a + b + c)
a
2
+ b
2
+ c
2
4
≥ 27abc(ab + bc + ca)
a
3
b + ab
3
,
✇❤✐❝❤ ✐s tr✉❡ ❜❡❝❛✉s❡
(a) (a + b + c)
a
2
+ b
2
+ c
2
≥ 9abc,
(b) a
2
+ b
2
+ c
2
≥ ab + bc + ca,
(c) 2
a
2
+ b
2
+ c
2
2
≥ 3
a
3
b + ab
3
,
✇❤✐❝❤
(c)
✷✵
✐s ❡q✉✐✈❛❧❡♥t t♦
a
2
− ab + b
2
(a − b)
2
≥ 0,
✇❤✐❝❤ ✐s tr✉❡✳
✷✼✱ ▲❡t
a, b, c
❜❡ ♥♦♥♥❡❣❛t✐✈❡ ♥✉♠❜❡rs✱ ♥♦ t✇♦ ♦❢ ✇❤✐❝❤ ❛r❡ ③❡r♦✳ Pr♦✈❡ t❤❛t
a
2
(b + c)
b
2
+ bc + c
2
+
b
2
(c + a)
c
2
+ ca + a
2
+
c
2
(a + b)
a
2
+ ab + b
2
2(a
2
+ b
2
+ c
2
)
a + b + c
.
❙♦❧✉t✐♦♥✿
a
2
(b + c)
b
2
+ bc + c
2
=
4a
2
(b + c)(ab + bc + ca)
(b
2
+ bc + c
2
) (ab + bc + ca)
≥
4a
2
(b + c)(ab + bc + ca)
(b
2
+ bc + c
2
+ ab + bc + ca)
2
=
4a
2
(ab + bc + ca)
(b + c)(a + b + c)
2
,
✐t s✉❢❢✐❝❡s t♦ ♣r♦✈❡
a
2
b + c
≥
(a + b + c)
a
2
+ b
2
+ c
2
2(ab + bc + ca)
,
♦r
a
2
b + c
+ a
≥
(a + b + c)
3
2(ab + bc + ca)
,
♦r
a
b + c
≥
(a + b + c)
2
2(ab + bc + ca)
,
✇❤✐❝❤ ✐s tr✉❡ ❜② ❈❛✉❝❤②✲❙❝❤✇❛r③ ✐♥❡q✉❛❧✐t②
a
b + c
=
a
2
a(b + c)
≥
(a + b + c)
2
2(ab + bc + ca)
.
❲❡ ❥✉st ✇❛♥t t♦ ❣✐✈❡ ❛ ❧✐tt❧❡ ♥♦t❡ ❤❡r❡✳ ◆♦t✐❝❡ t❤❛t
a
2
(b + c)
b
2
+ bc + c
2
+
a(b + c)
a + b + c
=
a(b + c)(a
2
+ b
2
+ c
2
+ ab + bc + ca)
(b
2
+ bc + c
2
)(a + b + c)
,
❛♥❞
2(a
2
+ b
2
+ c
2
)
a + b + c
+
a(b + c)
a + b + c
=
2(a
2
+ b
2
+ c
2
+ ab + bc + ca)
a + b + c
.
❚❤❡r❡❢♦r❡✱ t❤❡ ✐♥❡q✉❛❧✐t② ❝❛♥ ❜❡ ✇r✐tt❡♥ ✐♥ t❤❡ ❢♦r♠
a(b + c)
b
2
+ bc + c
2
+
b(c + a)
c
2
+ ca + a
2
+
c(a + b)
a
2
+ ab + b
2
≥ 2,
◆♦t❡ t❤❛t
❝②❝
a(b + c)
b
2
+ bc + c
2
=
❝②❝
4a(b + c)(ab + bc + ca)
4(b
2
+ bc + c
2
)(ab + bc + ca)
❝②❝
4a(ab + bc + ca)
(b + c)(a + b + c)
2
.
✷✶
❙♦ t❤❛t ❲❡ ❤❛✈❡ t♦ ♣r♦✈❡✿
❝②❝
4a(ab + bc + ca)
(b + c)(a + b + c)
2
2,
♦r
❝②❝
a
b + c
(a + b + c)
2
2(ab + bc + ca)
,
✇❤✐❝❤ ✐s ♦❜✈✐♦✉s❧② tr✉❡ ❞✉❡ t♦ t❤❡ ❈❛✉❝❤②✲❙❝❤✇❛r③ ✐♥❡q✉❛❧✐t②✳
❚❤✐s ✐s ❛♥♦t❤❡r ♥❡✇ ❙♦❧✉t✐♦♥✳ ❋✐rst✱ ❲❡ ✇✐❧❧ ♣r♦✈❡ t❤❛t
(a
2
+ ac + c
2
)(b
2
+ bc + c
2
) ≤
ab(a + b) + bc(b + c) + ca(c + a)
a + b
.(1)
✐♥❞❡❡❞✱ ✉s✐♥❣ t❤❡ ❈❛✉❝❤②✲❙❝❤✇❛r③ ✐♥❡q✉❛❧✐t②✱ ❲❡ ❤❛✈❡
√
ac ·
√
bc +
a
2
+ ac + c
2
·
b
2
+ bc + c
2
≤
(ac + a
2
+ ac + c
2
)(bc + b
2
+ bc + c
2
)
= (a + c)(b + c).
✐t ❢♦❧❧♦✇s t❤❛t
(a
2
+ ac + c
2
)(b
2
+ bc + c
2
) ≤ ab + c
2
+ c
a + b −
√
ab
≤ ab + c
2
+ c
a + b −
2ab
a + b
=
ab(a + b) + bc(b + c) + ca(c + a)
a + b
.
◆♦✇✱ ❢r♦♠ ✭✶✮✱ ✉s✐♥❣ t❤❡ ❆▼✲●▼ ✐♥❡q✉❛❧✐t②✱ ❲❡ ❣❡t
1
a
2
+ ac + c
2
+
1
b
2
+ bc + c
2
≥
2
(a
2
+ ac + c
2
)(b
2
+ bc + c
2
)
≥
2(a + b)
ab(a + b) + bc(b + c) + ca(c + a)
.
(2)
❋r♦♠
(2)
❲❡ ❤❛✈❡
a(b + c)
b
2
+ bc + c
2
=
ab
1
a
2
+ ac + c
2
+
1
b
2
+ bc + c
2
≥
2ab(a + b)
ab(a + b) + bc(b + c) + ca(c + a)
= 2.
✷✾✱ ✐❢
a, b, c > 0
t❤❡♥ t❤❡ ❢♦❧❧♦✇✐♥❣ ✐♥❡q✉❛❧✐t② ❤♦❧❞s✿
a
2
(b + c)
b
2
+ bc + c
2
+
b
2
(c + a)
c
2
+ ca + a
2
+
c
2
(a + b)
a
2
+ ab + b
2
≥ 2
a
3
+ b
3
+ c
3
a + b + c
❚❤✐s ✐♥❡q✉❛❧✐t② ✐s ❡q✉✐✈❛❧❡♥t t♦
a
2
(b + c)(a + b + c)
b
2
+ bc + c
2
≥ 2
(a
3
+ b
3
+ c
3
) (a + b + c)
♦r
a
2
+
a
2
(ab + bc + ca)
b
2
+ bc + c
2
≥ 2
(a
3
+ b
3
+ c
3
) (a + b + c),
✷✷
❜❡❝❛✉s❡
2
(a
3
+ b
3
+ c
3
) (a + b + c) ≤
a
2
+ b
2
+ c
2
+
a
3
+ b
3
+ c
3
(a + b + c)
a
2
+ b
2
+ c
2
,
✐t s✉❢❢✐❝❡s t♦ ♣r♦✈❡ t❤❛t
a
2
b
2
+ bc + c
2
≥
a
3
+ b
3
+ c
3
(a + b + c)
(a
2
+ b
2
+ c
2
) (ab + bc + ca)
,
❜② ❈❛✉❝❤②✲❙❝❤✇❛r③ ✐♥❡q✉❛❧✐t②✱ ❲❡ ❤❛✈❡
a
2
b
2
+ bc + c
2
≥
a
2
+ b
2
+ c
2
2
a
2
(b
2
+ bc + c
2
)
=
a
2
+ b
2
+ c
2
2
2
a
2
b
2
+
a
2
bc
,
✐t s✉❢❢✐❝❡s t♦ ♣r♦✈❡ t❤❛t
a
2
+ b
2
+ c
2
3
(ab + bc + ca) ≥
a
3
+ b
3
+ c
3
(a + b + c)
2
a
2
b
2
+
a
2
bc
.
▲❡t
A =
a
4
, B =
1
2
a
3
b + ab
3
, C =
a
2
b
2
, D =
a
2
bc,
❲❡ ❤❛✈❡
a
2
+ b
2
+ c
2
2
= A + 2C,
a
2
+ b
2
+ c
2
(ab + bc + ca) = 2B + D,
a
3
+ b
3
+ c
3
(a + b + c) = A + 2B,
❛♥❞
2
a
2
b
2
+
a
2
bc = 2C + D.
❚❤❡r❡❢♦r❡✱ ✐t s✉❢❢✐❝❡s t♦ ♣r♦✈❡ t❤❛t
(A + 2C) (2B + D) ≥ (A + 2B) (2C + D) ,
♦r
2 (A − D) (B − C) ≥ 0,
✇❤✐❝❤ ✐s tr✉❡ ❜❡❝❛✉s❡
A ≥ D
❛♥❞
B ≥ C
✸✵✱ ●✐✈❡♥
a, b, c ≥ 0
s✉❝❤ t❤❛t
a + b + c = 1
Pr♦✈❡ t❤❛t
2
a
2
b + b
2
c + c
2
a + ab + bc + ca ≤ 1
❘❡✇r✐t❡ t❤❡ ✐♥❢♦r♠ ✐♥❡q✉❛❧✐t② ❛s
2
a
2
b + b
2
c + c
2
a + ab + bc + ca ≤ (a + b + c)
2
✷✸
2
(a
2
b + b
2
c + c
2
a) (a + b + c) ≤ a
2
+ b
2
+ c
2
+ ab + bc + ca
❆ss✉♠❡ t❤❛t ❜ ✐s t❤❡ ♥✉♠❜❡r ❜❡t✐❡♥ ❛ ❛♥❞ ❝✳ ❚❤❡♥✱ ❜② ❛♣♣❧②✐♥❣ t❤❡ ❆▼✲●▼ ✐♥❡q✉❛❧✐t②✱ ❲❡
❣❡t
2
(a
2
b + b
2
c + c
2
a) (a + b + c) ≤
a
2
b + b
2
c + c
2
a
b
+ b(a + b + c)
✐t ✐s t❤✉s s✉❢❢✐❝✐❡♥t t♦ ♣r♦✈❡ t❤❡ str♦♥❣❡r ✐♥❡q✉❛❧✐t②
a
2
+ b
2
+ c
2
+ ab + bc + ca ≥
a
2
b + b
2
c + c
2
a
b
+ b(a + b + c)
❚❤✐s ✐♥❡q✉❛❧✐t② ✐s ❡q✉✐✈❛❧❡♥t t♦
c(a − b)(b − c)
b
≥ 0,
✇❤✐❝❤ ✐s ♦❜✈✐♦✉s❧② tr✉❡ ❛❝❝♦r❞✐♥❣ t♦ t❤❡ ❛ss✉♠♣t✐♦♥ ♦❢
b
❍♦✇ t♦ ♣r♦✈❡
a
4
+ 2
a
3
c ≥
a
2
b
2
+ 2
a
3
b
♦♥❧② ❜② ❆▼✲●▼ ❊q✉✐✈❛❧❡♥t t♦ ♣r♦✈❡
(a − b)
2
(a + b)
2
≥ 4(a − b)(b − c)(a − c)(a + b + c)
❲▲❖● ❲❡ ❝❛♥ ❛ss✉♠❡ t❤❛t
a ≥ b ≥ c, a − b = x, b − c = y
t❤❡♥ ❲❡ ♥❡❡❞ t♦ ♣r♦✈❡ t❤❛t
x
2
(2c + 2y + x)
2
+ y
2
(2c + y)
2
+ (x + y)
2
(2c + x + y)
2
≥ xy(x + y)(3c + 2x + y)
❜②
(x + y)
4
≥ xy(x + y)(x + 2y)
❛♥❞
(x + y)
3
≥ 3xy(x + y)
❲❡ ❤❛✈❡ ❝♦♠♣❧❡t❡❞ t❤❡ ❙♦❧✉t✐♦♥
✸✶✱ ▲❡t
a, b, c
❜❡ ♣♦s✐t✐✈❡ ♥✉♠❜❡rs s✉❝❤ t❤❛t
a
2
b
2
+ b
2
c
2
+ c
2
a
2
≥ a
2
b
2
c
2
❋✐♥❞ t❤❡ ♠✐♥✐♠✉♠ ♦❢ ❆
A =
a
2
b
2
c
3
(a
2
+ b
2
)
+
b
2
c
2
a
3
(b
2
+ c
2
)
+
c
2
a
2
b
3
(c
2
+ a
2
)
◆♦ ♦♥❡ ❧✐❦❡ t❤✐s ♣r♦❜❧❡♠❄ ❙❡tt✐♥❣
x =
1
a
, y =
1
b
, z =
1
c
✷✹
❲❡ ❤❛✈❡
x
2
+ y
2
+ z
2
≥ 1
❲❡ ✇✐❧❧ ♣r♦✈❡ t❤❛t
x
3
y
2
+ z
2
+
y
3
x
2
+ z
2
+
z
3
x
2
+ y
2
≥
√
3
2
❯s✐♥❣ ❈❛✉❝❤②✲❙❝❤✇❛r③✿
LHS ≥
(x
2
+ y
2
+ z
2
)
2
x(y
2
+ z
2
) + y(x
2
+ z
2
) + z(x
2
+ y
2
)
❇② ❆▼✲●▼ ❲❡ ❤❛✈❡✿
x(y
2
+z
2
)+y(x
2
+z
2
)+z(x
2
+y
2
) ≤
2
3
(x
2
+y
2
+z
2
)(x+y+z) ≤
2
√
3
(x
2
+y
2
+z
2
)
x
2
+ y
2
+ z
2
❇❡❝❛✉s❡
x
2
+ y
2
+ z
2
≥ 1
❙♦
(x
2
+ y
2
+ z
2
)
2
2
√
3
(x
2
+ y
2
+ z
2
)
x
2
+ y
2
+ z
2
≥
√
3
2
❲❡ ❞♦♥❡✦
✸✷✳
▲❡t ①✱②✱③ ❜❡ ♥♦♥ ♥❡❣❛t✐✈❡ r❡❛❧ ♥✉♠❜❡rs s✉❝❤ t❤❛t x
2
+ y
2
+ z
2
= 1
✳ ❢✐♥❞ t❤❡ ♠✐♥✐♠✉♠ ❛♥❞ ♠❛①✐♠✉♠ ♦❢ f = x + y + z − xyz.
❙♦❧✉t✐♦♥ ✶✳
❋✐rst ❲❡ ❢✐① ③ ❛♥❞ ❧❡t m = x+y = x+
√
1 − x
2
− z
2
= g(x)(0 ≤ x ≤
√
1 − z
2
), t❤❡♥ ❲❡ ❤❛✈❡
g
(x) = 1 −
x
√
1 − x
2
− z
2
,
❲❡ ❣❡t
g
(x) > 0 ⇔ 0 ≤ x <
1 − z
2
2
❛♥❞
g
(x) < 0 ⇔
1 − z
2
2
< x ≤
1 − z
2
,
s♦ ❲❡ ❤❛✈❡
m
min
= min{g(0), g(
1 − z
2
)} =
1 − z
2
❛♥❞
m
max
= g
1 − z
2
2
=
2 − 2z
2
.
❆❝t✉❛❧❧②✱ ❢ ❛♥❞ ✇r✐tt❡♥ ❛s
f = f(m) = −
z
2
m
2
+ m +
1 − z
2
z
2
+ z,
❡❛s② t♦ ♣r♦✈❡ t❤❛t t❤❡ ❛①✐s ♦❢ s②♠♠❡tr②
m =
1
z
>
2 − 2z
2
✷✺
