Bất đẳng thức đại số trong tam giác

Số hóa bởi trung tâm học liệu />Số hóa bởi trung tâm học liệu />Số hóa bởi trung tâm học liệu />Số hóa bởi trung tâm học liệu />Số hóa bởi trung tâm học liệu />•
•
•
•
•

sym
sym symmetric.
•

cyc
cyc cyclic.
• m
a
; m
b
; m
c
A, B, C.
• l
a
; l
b
; l
c
A, B, C.
• h
a
; h
b
; h

c
A, B, C.
• r :
• R :
• r
a
; r
b
; r
c
Số hóa bởi trung tâm học liệu />• p
• S
Số hóa bởi trung tâm học liệu />ABC AB = c; AC = b; BC = a
ABC
a
sin A
=
b
sin B
=
c
sin C
= 2R.
ABC
a
2
= b
2
+ c
2

− 2bc cos A;
b
2
= a
2
+ c
2
− 2ac cos B;
c
2
= a
2
+ b
2
− 2ab cos C.
ABC
a − b
a + b
=
tan
A − B
2
tan
A + B
2
= tan
A − B
2
tan
C

2
;
Số hóa bởi trung tâm học liệu />b − c
b + c
=
tan
B −C
2
tan
B + C
2
= tan
B −C
2
tan
A
2
;
c − a
c + a
=
tan
C − A
2
tan
C + A
2
= tan
C − A
2

tan
B
2
.
h
a
=
2S
a
=
2

p(p − a)(p − b)(p − c)
a
;
h
b
=
2S
b
=
2

p(p − a)(p − b)(p − c)
b
;
h
c
=
2S

c
=
2

p(p − a)(p − b)(p − c)
c
.
m
2
a
=
b
2
+ c
2
2
−
a
2
4
;
m
2
b
=
a
2
+ c
2
2

−
b
2
4
;
m
2
c
=
a
2
+ b
2
2
−
c
2
4
.
l
a
=
2bc
b + c
cos
A
2
;
l
b

=
2ac
a + c
cos
B
2
;
l
c
=
2ab
a + b
cos
C
2
.
R =
a
2 sin A
=
b
2 sin B
=
c
2 sin C
=
abc
4S
=
abc

4

p(p − a)(p − b)(p − c)
.
Số hóa bởi trung tâm học liệu />r = (p−a) tan
A
2
= (p−b) tan
B
2
= (p−c) tan
C
2
=
S
p
=

(p − a)(p − b)(p − c)
p
.
r
a
= p tan
A
2
=
S
p − a
=


p(p − b)(p − c)
p − a
;
r
b
= p tan
B
2
=
S
p − b
=

p(p − c)(p − a)
p − b
;
r
c
= p tan
C
2
=
S
p − c
=

p(p − a)(p − b)
p − c
.

S =
1
2
ah
a
=
1
2
bh
b
=
1
2
ch
c
=
1
2
bc sin A =
1
2
casinB =
1
2
ab sin C
= pr =
abc
4R
= (p − a)r
a

= (p − b)r
b
= (p − c)r
c
=

p(p − a)(p − b)(p − c)
=
√
rr
a
r
b
r
c
= p
2
tan
A
2
tan
B
2
tan
C
2
= 2R
2
sin A sin B sin C
=

1
4

2(a
2
b
2
+ b
2
c
2
+ c
2
a
2
) − (a
4
+ b
4
+ c
4
).
a = b cos C + c cos B = r(cot
B
2
+ cot
C
2
);
b = c cos A + a cos C = r(cot

C
2
+ cot
A
2
);
Số hóa bởi trung tâm học liệu />c = a cos B + b cos A = r(cot
A
2
+ cot
B
2
).
∆ABC BC = a, CA = b, AB = c
x
3
− 2px
2
+ (p
2
+ r
2
+ 4Rr)x − 4Rrp = 0.
tan
A
2
=
r
p − a
, a = 2R sin A,

a = 2R
2 tan
A
2
1 + tan
2
A
2
a = 4R
r
p − a
1 + (
r
p − a
)
2
= 4Rr
p − a
r
2
+ (p − a)
2
.
a(a
2
− 2pa + p
2
+ r
2
) = 4Rr(p − a),

a
3
− 2pa
2
+ (p
2
+ r
2
+ 4Rr)a − 4Rrp = 0.
a
x
3
− 2px
2
+ (p
2
+ r
2
+ 4Rr)x − 4Rrp = 0.
b c
Số hóa bởi trung tâm học liệu />∆ABC BC = a, CA = b, AB = c
p r R
a + b + c = 2p.
ab + bc + ca = p
2
+ r
2
+ 4Rr.
abc = 4Rrp.
x

3
− 2px
2
+ (p
2
+ r
2
+ 4Rr)x − 4Rrp = 0,
∆ABC BC = a, CA = b, AB = c
(a
2
+ 2Rr)(b
2
+ 2Rr)(c
2
+ 2Rr) = 2Rr(ab + bc + ca − 2Rr)
2
p = (a
2
+ 2Rr)(b
2
+ 2Rr) + (b
2
+ 2Rr)(c
2
+ 2Rr) + (c
2
+ 2Rr)(a
2
+ 2Rr)

= (p
2
+ 2Rr + r
2
)
2
− 2Rr.
a, b, c
x
3
− 2px
2
+ (p
2
+ 4Rr + r
2
)x − 4Rrp = 0.
y = x
2
+ 2Rr T = p
2
+ 2Rr + r
2
. x
y
y
3
− (4p
2
+ 2Rr − 2T )y

2
+ (T
2
− 2Rr)y − 2RrT
2
= 0.
y
1
, y
2
, y
3
ab +bc+ca = T +2Rr
y
1
y
2
y
3
= 2RrT
2
= 2Rr(ab + bc + ca − 2Rr)
2
∆ABC BC = a, CA = b, AB = c
r
1
r
2
r
3

S =

(a
2
+ p
2
+ 4Rr + r
2
)(b
2
+ p
2
+ 4Rr + r
2
)(c
2
+ p
2
+ 4Rr + r
2
)
r
1
+ r
2
+ r
3
+ r
.
a, b, c

x
3
− 2px
2
+ (p
2
+ 4Rr + r
2
)x − 4Rrp = 0.
Số hóa bởi trung tâm học liệu />y = x
2
+ p
2
+ 4Rr + r
2
y
1
= a
2
+ p
2
+ 4Rr + r
2
,
y
2
= b
2
+ p
2

+ 4Rr + r
2
,
y
3
= c
2
+ p
2
+ 4Rr + r
2
x

x
3
− 2px
2
+ (p
2
+ 4Rr + r
2
)x − 4Rrp = 0
x
2
+ p
2
+ 4Rr + r
2
− y = 0.


x
3
− 2px
2
+ (p
2
+ 4Rr + r
2
)x − 4Rrp = 0
x
3
+ p
2
x + 4Rrx + r
2
x − yx = 0.
2px
2
−yx +4Rrp = 0 2p

y − p
2
− 4Rr − r
2

−yx +4Rrp = 0.
x =
2py − 4Rrp − 2r
2
p

y
.

2py − 4Rrp − 2r
2
p
y

2
+ p
2
+ 4Rr + r
2
= y,
y
3
− (p
2
+ 4Rr + r
2
)y
2
− (2py − 4Rrp − 2r
2
p)
2
= 0.
y
1
, y

2
, y
3
y
1
y
2
y
3
= (4R + 2r)
2
r
2
p
2
.
(a
2
+p
2
+4Rr +r
2
)(b
2
+p
2
+4Rr +r
2
)(c
2

+p
2
+4Rr +r
2
) = (4Rr + 2r
2
)
2
p
2
.
r
1
+ r
2
+ r
3
= 4R + r
∆ABC
h
a
, h
b
, h
c
h
a
, h
b
, h

c
y
3
−
S
2
+ 4Rr
3
+ r
4
2Rr
2
y
2
+
2S
2
Rr
y −
2S
2
R
= 0.
Số hóa bởi trung tâm học liệu />h
a
+ h
b
+ h
c
=

S
2
+ 4Rr
3
+ r
4
2Rr
2
.
h
a
h
b
+ h
b
h
c
+ h
c
h
a
=
2S
2
Rr
h
a
h
b
h

c
=
2S
2
R
.
(h
a
− r)(h
b
− r)(h
c
− r) =
S
2
+ 2Rr
3
+ r
4
2R
.
a, b, c
x
3
− 2px
2
+ (p
2
+ 4Rr + r
2

)x − 4Rrp = 0
2S
a
,
2S
b
,
2S
c
2S
2
−
2S
2
r
y +
S
2
r
2
+ 4Rr + r
2
2
y
2
− Ry
3
= 0.
h
a

, h
b
, h
c
y
3
−
S
2
+ 4Rr
3
+ r
4
2Rr
2
y
2
+
2S
2
Rr
y −
2S
2
R
= 0.
y
3
−
S

2
+ 4Rr
3
+ r
4
2Rr
2
y
2
+
2S
2
Rr
y −
2S
2
R
= (y − h
a
)(y − h
b
)(y − h
c
).
y = r
(h
a
− r)(h
b
− r)(h

c
− r) =
S
2
+ 2Rr
3
+ r
4
2R
.
∆ABC
r
1
, r
2
, r
3
r
1
, r
2
, r
3
x
3
− (4R + r)x
2
+ p
2
x − p

2
r = 0.
r
1
+ r
2
+ r
3
= 4R + r
r
1
r
2
+ r
2
r
3
+ r
3
r
1
= p
2
.
Số hóa bởi trung tâm học liệu />r
1
r
2
r
3

= p
2
r.
(r
1
− r)(r
2
− r)(r
3
− r) = 4Rr
2
.
tan
A
2
=
r
1
p
, a = 2R sin A
a = 2R
2 tan
A
2
1 + tan
2
A
2
,
a = 4R

r
1
p
1 +
r
2
1
p
2
= 4Rr
1
p
r
2
1
+ p
2
.
r
1
(p − a) = S = rp
(r
1
− r)p
r
1
= a = 4Rr
1
p
r

2
1
+ p
2
,
(r
1
− r)(r
2
1
+ p
2
) = 4Rr
2
1
.
r
1
x
3
−(4R + r)x
2
+ p
2
x −p
2
r = 0 r
2
r
3

x
3
−(4R + r)x
2
+ p
2
x −p
2
r = 0
x
3
− (4R + r)x
2
+ p
2
x − p
2
r = (x − r
1
)(x − r
2
)(x − r
3
),
(r
1
− r)(r
2
− r)(r
3

− r) = 4Rr
2
.
∆ABC h
a
, h
b
, h
c
,
r
1
, r
2
, r
3
Số hóa bởi trung tâm học liệu />4

1
h
2
a
+
1
h
2
b
+
1
h

2
c

=
1
r
2
1
+
1
r
2
2
+
1
r
2
3
+
1
r
2
h
a
h
b
+ h
b
h
c

+ h
c
h
a
=
2r
R
(r
1
r
2
+ r
2
r
3
+ r
3
r
1
).
r
1
, r
2
, r
3
x
3
−(4R+r)x
2

+p
2
x−p
2
r = 0
1
r
1
,
1
r
2
,
1
r
3
p
2
rx
3
− p
2
x
2
+ (4R + r)x − 1 = 0.
1
r
2
1
+

1
r
2
2
+
1
r
2
3
=
1
r
2
− 2
4R + r
p
2
r
.
a, b, c
x
3
− 2px
2
+ (p
2
+ r
2
+ 4Rr)x − 4Rrp = 0
a

2
+ b
2
+ c
2
= 4p
2
− 2(p
2
+ r
2
+ 4Rr) = 2p
2
− 2r
2
− 8Rr
4

1
h
2
a
+
1
h
2
b
+
1
h

2
c

=
2p
2
− 2r
2
− 8Rr
p
2
r
2
=
2
r
2
− 2
4R + r
p
2
r
.
4

1
h
2
a
+

1
h
2
b
+
1
h
2
c

=
1
r
2
1
+
1
r
2
2
+
1
r
2
3
+
1
r
2
.

h
a
h
b
+ h
b
h
c
+ h
c
h
a
= r
2S
2
Rr
2
= 2r
p
2
R
=
2r
R
(r
1
r
2
+ r
2

r
3
+ r
3
r
1
).
∆ABC
r
1
, r
2
, r
3

r
1
r
− 1

r
2
r
− 1

r
3
r
− 1


= 4
R
r
d
2
a
+ d
2
b
+ d
2
c
= 11R
2
+ 2Rr, d
a
, d
b
, d
c
∆ABC
Số hóa bởi trung tâm học liệu />x
3
− (4R + r)x
2
+ p
2
x − p
2
r = (x − r

1
)(x − r
2
)(x − r
3
)
x = r (r
1
− r)(r
2
− r)(r
3
− r) = 4Rr
2
r
3

r
1
r
− 1

r
2
r
− 1

r
3
r

− 1

= 4
R
r
.
d
2
a
= R
2
+ 2Rr
1
, d
2
b
= R
2
+ 2Rr
2
d
2
c
= R
2
+ 2Rr
3
d
2
a

+ d
2
b
+ d
2
c
= 11R
2
+ 2Rr.
∆ABC
r
1
, r
2
, r
3
; r
2.
r
1
− r
r
1
+ r
.
r
2
− r
r
2

+ r
.
r
3
− r
r
3
+ r
+
r
1
− r
r
1
+ r
.
r
2
− r
r
2
+ r
+
r
2
− r
r
2
+ r
.

r
3
− r
r
3
+ r
+
r
3
− r
r
3
+ r
.
r
1
− r
r
1
+ r
= 1.
r
1
, r
2
, r
3
x
3
− (4R + r)x

2
+ p
2
x − p
2
r = 0 (1).
y =
x − r
x + r
y = 1 x =
r(y + 1)
1 − y
x
(r
2
+ 2Rr + p
2
)y
3
+ (2r
2
+ 2Rr − 2p
2
)y
2
+ (r
2
− 2Rr + p
2
)y − 2Rr = 0.

y
1
, y
2
, y
3
2y
1
y
2
y
3
+ y
1
y
2
+ y
2
y
3
+ y
3
y
1
=
4Rr + r
2
− 2Rr + p
2
r

2
+ 2Rr + p
2
= 1.
∆ABC r
1
, r
2
, r
3
; r
T
1
=
2r
1
− r
r
1
+ r
+
2r
2
− r
r
2
+ r
+
2r
3

− r
r
3
+ r
,
T
2
=
2r
1
− r
r
1
+ r
.
2r
2
− r
r
2
+ r
+
2r
2
− r
r
2
+ r
.
2r

3
− r
r
3
+ r
+
2r
3
− r
r
3
+ r
.
2r
1
− r
r
1
+ r
,
T
3
=
2r
1
− r
r
1
+ r
.

2r
2
− r
r
2
+ r
.
2r
3
− r
r
3
+ r
.
Số hóa bởi trung tâm học liệu />4T
3
+ T
2
− 2T
1
= 5.
r
1
, r
2
, r
3
x
3
− (4R + r)x

2
+ p
2
x − p
2
r = 0, (1).
y =
2x − r
x + r
y = 2 x =
r(y + 1)
2 − y
x
2(r
2
+ 2Rr + p
2
)y
3
+ 3(r
2
−3p
2
)y
2
−2(6Rr −6p
2
)y −(r
2
+ 8Rr + 4p

2
) = 0.
y
1
, y
2
, y
3
2T
1
= 2(y
1
+ y
2
+ y
3
) =
9p
2
− 3r
2
r
2
+ 2Rr + p
2
,
T
2
= y
1

y
2
+ y
2
y
3
+ y
3
y
1
=
6p
2
− 6Rr
r
2
+ 2Rr + p
2
,
4T
3
= 4y
1
y
2
y
3
=
2r
2

+ 16Rr + 8p
2
r
2
+ 2Rr + p
2
.
4T
3
+ T
2
− 2T
1
= 5
∆ABC a, b, c; p
r
(a − b)
2
ab
+
(b − c)
2
bc
+
(c − a)
2
ca
=
p
2

+ r
2
2Rr
− 7.
x
1
, x
2
, x
3
x
3
+a
1
x
2
+a
2
x+a
3
= 0
(x
2
1
− x
2
x
3
)(x
2

2
− x
3
x
1
)(x
2
3
− x
1
x
2
) = a
3
1
a
3
− a
3
2
.
(x
1
− x
2
)
2
x
1
x

2
+
(x
2
− x
3
)
2
x
2
x
3
+
(x
3
− x
1
)
2
x
3
x
1
=
a
1
a
2
a
3

− 9.
a, b, c
x
3
− 2px
2
+ (p
2
+ r
2
+ 4Rr)x − 4Rrp = 0
Số hóa bởi trung tâm học liệu />(a
2
− bc)(b
2
− ca)(c
2
− ab) = 32p
4
Rr − (p
2
+ r
2
+ 4Rr)
3
.
(a − b)
2
ab
+

(b − c)
2
bc
+
(c − a)
2
ca
=
2p(p
2
+ r
2
+ 4Rr)
4Rrp
− 9.
(a − b)
2
ab
+
(b − c)
2
bc
+
(c − a)
2
ca
=
p
2
+ r

2
2Rr
− 7.
∆ABC a, b, c
r, R
r
1
, r
2
, r
3
p S
(r
1
− r
2
)
2
r
1
r
2
+
(r
2
− r
3
)
2
r

2
r
3
+
(r
3
− r
1
)
2
r
3
r
1
=
4R
r
− 8.
r
1
, r
2
, r
3
x
3
−(4R+r)x
2
+p
2

x−p
2
r = 0
(r
1
− r
2
)
2
r
1
r
2
+
(r
2
− r
3
)
2
r
2
r
3
+
(r
3
− r
1
)

2
r
3
r
1
=
(4R + r)p
2
p
2
r
− 9 =
4R
r
− 8.
∆ABC R
r
1
, r
2
, r
3
p

r
2
1
p
2
+ 1


r
2
2
p
2
+ 1

r
2
3
p
2
+ 1

=
16R
2
p
2
.
r
1
, r
2
, r
3
x
3
− (4R + r)x

2
+ p
2
x − p
2
r = 0
y
1
= r
2
1
+ p
2
, y
2
= r
2
2
+ p
2
, y
3
= r
2
3
+ p
2
x

x

3
− (4R + r)x
2
+ p
2
x − p
2
r = 0
x
2
+ p
2
− y = 0.

x
3
− (4R + r)x
2
+ p
2
x − p
2
r = 0
x
3
+ p
2
x − yx = 0.
Số hóa bởi trung tâm học liệu />(4R + r)x
2

− yx + p
2
r = 0;
(4R + r)(y − p
2
) − yx + p
2
r = 0.
T = 4R + r x =
T y − 4Rp
2
y
(T y − 4Rp
2
)
2
y
2
+ p
2
− y = 0.
y
3
− (T
2
+ p
2
)y
2
+ 8RT p

2
y − 16R
2
p
4
= 0.
y
1
, y
2
, y
3
(r
2
1
+ p
2
)(r
2
2
+ p
2
)(r
2
3
+ p
2
)
p
6

=
y
1
y
2
y
3
p
6
=
16R
2
p
4
p
6
.

r
2
1
p
2
+ 1

r
2
2
p
2

+ 1

r
2
3
p
2
+ 1

=
16R
2
p
2
.
∆ABC R
r r
1
, r
2
, r
3
p
(i)

r
2
1
p
2

− 1

r
2
2
p
2
− 1

r
2
3
p
2
− 1

= 4

(2R + r)
2
p
2
− 1

.
(ii)

r
2
1

p
2
− 1

r
2
2
p
2
− 1

+

r
2
2
p
2
− 1

r
2
3
p
2
− 1

+

r

2
3
p
2
− 1

r
2
1
p
2
− 1

= 8 − 4
(2R + r)(r
1
+ r
2
+ r
3
)
p
2
.
r
1
, r
2
, r
3

x
3
− (4R + r)x
2
+ p
2
x − p
2
r = 0
y
1
= r
2
1
− p
2
, y
2
= r
2
2
− p
2
, y
3
= r
2
3
− p
2

x

x
3
− (4R + r)x
2
+ p
2
x − p
2
r = 0
x
2
− p
2
− y = 0.
Số hóa bởi trung tâm học liệu />
x
3
− (4R + r)x
2
+ p
2
x − p
2
r = 0
x
3
− p
2

x − yx = 0.
(4R + r)x
2
− yx − 2p
2
x + p
2
r = 0.
(4R + r)(y + p
2
) − (y + 2p
2
)x + p
2
r = 0.
T = 4R + r x =
T y + (T + r)p
2
y + p
2
(T y + (T + r)p
2
)
2
(y + 2p
2
)
2
− p
2

− y = 0.
y(y + 2p
2
)
2
+ p
2
(y + 2p
2
)
2
− (T y + (T + r)p
2
)
2
= 0.
y
1
, y
2
, y
3
(r
2
1
− p
2
)(r
2
2

− p
2
)(r
2
3
− p
2
)
p
6
=
y
1
y
2
y
3
p
6
=
(4R + 2r)
2
p
4
− 4p
6
p
6
.


r
2
1
p
2
− 1

r
2
2
p
2
− 1

r
2
3
p
2
− 1

= 4

(2R + r)
2
p
2
− 1

.

y 8p
4
− 4(2R + r)(4R + r)p
2
.
Số hóa bởi trung tâm học liệu />a, b, c
1
a
3
+ b
3
+ abc
+
1
b
3
+ c
3
+ abc
+
1
c
3
+ a
3
+ abc
≤
1
abc
(a − b)(a

2
− b
2
) ≥ 0, a
3
+ b
3
≥ ab(a + b).
1
a
3
+ b
3
+ abc
≤
1
ab(a + b) + abc
=
c
abc(a + b + c)
.
1
b
3
+ c
3
+ abc
≤
a
abc(a + b + c)

1
c
3
+ a
3
+ abc
≤
b
abc(a + b + c)
.
1
a
3
+ b
3
+ abc
+
1
b
3
+ c
3
+ abc
+
1
c
3
+ a
3
+ abc

≤
1
abc
.
a = b = c.
Số hóa bởi trung tâm học liệu />∆ABC
1
R
2
≤
1
bc
+
1
ca
+
1
ab
.
1
R
2
≤
1
a
2
+
1
b
2

+
1
c
2
≤
1
4r
2
.

a

b

c
abc
≥
64S
3
(p
2
+ r
2
+ 4Rr − bc)(p
2
+ r
2
+ 4Rr − ca)(p
2
+ r

2
+ 4Rr − ab)
.
1
a
3
+ b
3
+ 4RS
+
1
b
3
+ c
3
+ 4RS
+
1
c
3
+ a
3
+ 4RS
≤
1
4r
2
(a + b + c)
.
a, b, c

x
3
− 2px
2
+ (p
2
+ r
2
+ 4Rr)x − 4Rrp = 0.
1
a
,
1
b
,
1
c
−1 + 2px − (p
2
+ r
2
+ 4Rr)x
2
+ 4Rrpx
3
= 0,
1
bc
+
1

ca
+
1
ab
=
2p
4Rrp
=
1
2Rr
≥
1
R
2
.
f(x) = x
3
−2px
2
+ (p
2
+ r
2
+ 4Rr)x −4Rrp = (x −a)(x−b)(x−c).
f (x) 3x −2p = (x−a)+(x−b) +(x−c)
3x − 2p
f(x)
=
1
(x − a)(x − b)

+
1
(x − b)(x − c)
+
1
(x − c)(x − a)
.
x = p,
1
(p − a)(p − b)
+
1
(p − b)(p − c)
+
1
(p − c)(p − a)
=
p
r
2
p
=
1
r
2
.
1
4r
2
=

1
4(p − a)(p − b)
+
1
4(p − b)(p − c)
+
1
4(p − c)(p − a)
≥
1
a
2
+
1
b
2
+
1
c
2
.
1
R
2
≤
1
ab
+
1
bc

+
1
ca
≤
1
a
2
+
1
b
2
+
1
c
2
.
Số hóa bởi trung tâm học liệu />1
R
2
≤
1
a
2
+
1
b
2
+
1
c

2
≤
1
4r
2
.

a
=
2

bc(p − a)p
b + c
, 
b
=
2

ca(p − b)p
c + a
, 
c
=
2

ab(p − c)p
a + b
.

a


b

c
=
32Rrabcp
2
S
(p
2
+ r
2
+ 4Rr − bc)(p
2
+ r
2
+ 4Rr − ca)(p
2
+ r
2
+ 4Rr − ab)
.
R ≥ 2r

a

b

c
abc

≥
64S
3
(p
2
+ r
2
+ 4Rr − bc)(p
2
+ r
2
+ 4Rr − ca)(p
2
+ r
2
+ 4Rr − ab)
.
1
a
3
+ b
3
+ 4RS
+
1
b
3
+ c
3
+ 4RS

+
1
c
3
+ a
3
+ 4RS
≤
1
4Rrp
.
1
4Rrp
≤
1
4r
2
(a + b + c)
,
1
a
3
+ b
3
+ 4RS
+
1
b
3
+ c

3
+ 4RS
+
1
c
3
+ a
3
+ 4RS
≤
1
4r
2
(a + b + c)
.
∆ABC
S
2
+ 5r
4
2R
≤ (h
a
− r)(h
b
− r)(h
c
− r) ≤
S
2

+ 5r
4
4r
1
h
3
a
+ h
3
b
+
2S
2
R
+
1
h
3
b
+ h
3
c
+
2S
2
R
+
1
h
3

c
+ h
3
a
+
2S
2
R
≤
1
h
a
h
b
h
c
a, b, c
x
3
− 2px
2
+ (p
2
+ 4Rr + r
2
)x − 4Rrp = 0
2S
a
,
2S

b
,
2S
c
2S
2
−
2S
2
r
y +
S
2
r
2
+ 4Rr + r
2
2
y
2
− Ry
3
= 0
h
a
, h
b
, h
c
Số hóa bởi trung tâm học liệu />y

3
−
S
2
+ 4Rr
3
+ r
4
2Rr
2
y
2
+
2S
2
Rr
y −
2S
2
R
= 0
y
3
−
S
2
+ 4Rr
3
+ r
4

2Rr
2
y
2
+
2S
2
Rr
y −
2S
2
R
= (y − h
a
)(y − h
b
)(y − h
c
)
y = r (h
a
− r)(h
b
− r)(h
c
− r) =
S
2
+ 2Rr
3

+ r
4
2R
R ≥ r
S
2
+ 5r
4
2R
≤ (h
a
− r)(h
b
− r)(h
c
− r) ≤
S
2
+ 5r
4
4r
.
∆ABC
8r
3
≤ (r
1
− r)(r
2
− r)(r

3
− r) ≤ R
3
.
81r
2
≤ (r
2
1
+ r
2
2
+ r
2
3
+ 2
r
1
r
2
r
3
r
≤
81
4
R
2
.
ab + bc + ca ≥ 4r(r

1
+ r
2
+ r
3
).
729r
3
− 3R(a + b + c)
2
≤ r
3
1
+ r
3
2
+ r
3
3
≤
729
8
R
3
− 6r(a + b + c)
2
.
1
r
3

1
+ r
3
2
+ Sp
+
1
r
3
2
+ r
3
3
+ Sp
+
1
r
3
3
+ r
3
1
+ Sp
≤
1
r
1
r
2
r

3
.
(r
1
− r)(r
2
− r)(r
3
− r) = 4Rr
2
R ≥ 2r
8r
3
≤ (r
1
− r)(r
2
− r)(r
3
− r) ≤ R
3
.
r
2
1
+ r
2
2
+ r
2

3
= (r
1
+ r
2
+ r
3
)
2
− 2(r
1
r
2
+ r
2
r
3
+ r
3
r
1
)
r
2
1
+ r
2
2
+ r
2

3
= (4R + r)
2
− 2
r
1
r
2
r
3
r
r
2
1
+ r
2
2
+ r
2
3
+ 2
r
1
r
2
r
3
r
= (4R + r)
2

Số hóa bởi trung tâm học liệu />