Các bài toán tự sáng tác

If are non-negative numbers, no two of which are zero, then
Loi giai :Nguyen Dinh Thanh Cong .
Nhan 2 ve voi 2(a+b+c) .
Chu y bdt sau :
Recalling the known inequality:
Ta chi can CM :
Ta dung BCS va xet 2 TH de CM BDT nay .
Ket thuc CM .
Ta con the su dung SOS de giai bai toan nay .
Bai Toan : Cho tam giác ABC , T là điểm Torixeli ( T thuộc miền trong tam giác ) .
AT,BT,CT cắt BC,CA,AB tại . là điểm đẳng giác của T qua
BC,CA,AB.CMR : đồng quy tại 1 điểm .
Loi giai , goi T’ la diem dang giac cua T . Ta doi xung voi T qua BC . Ta Cm Ta thuoc
AT de suy ra T’ thuoc AT1 .
CM su dung luong giac , dinh ly Sin .
Bai toan :Cho x,y thuoc [2009’-2009] thoa man (x^2-Y62-2xy)^2=4 .
Tim gia tri nho nhat cua x^2+y^2 .
Goi y :Su dung pt pell .
Lá thôi bay trong cơn mưa mùa hạ
Chút tinh khôi để lại còn vẹn nguyên
Huyền nhung thoáng ngây thơ 1 cơn gió
Hồ mơ lặng bóng dáng buồn xa xăm.
Bai toan: Given triangle . Let be an arbitray point on A-altitude.
,
.
Prove that the midpoints of and are collinear.
Loi giai : is collinear
from
True.
Done.
Bai toan : ABC is an acute-angled triangle. The incircle (I) touches BC at K. The altitude

AD has midpoint M. The line KM meets the incircle again at N.
. Show that
Loi giai : NK is the bisector of angle BNC, hence
Then
Cach khac: Let be the points of tangent of and and . It
is well-known that . Let be the midpoint of . Since is the polar
of with respect to , , thus , which yields and
are similar. But are respectively the median of and so
and are also similar. By this, we have . However, is the
tangent of so either is . By Newton's identity in a harmonic division we have
. This means is also the tangent of the circumcircle of
at . Therefore, the triangles and are similar, form the ratios,
we deduce that . Finally, so by Mc Laurin's identity, take
the geometrical length, we have and ,
divides those two equations we deduce that . Hence
, which means that , id est is the angle bisector
of . So . By this, . Q.E.D.
Bai toan: Let be the set of all positive integers
solve the following equation in :
Loi giai:Tran Nam Dung
WLOG we can assume that . We are to prove that . Suppose the
contradiction that . Consider cases:
1. If then . Hence
must not be a perfect square , contradiction.
2. If , notice that must be odd, since which and
are consecutive integers. Rewrite the equation in the following form:
We have so either or is divisible by . This
yields .
Thus, we can deduce that:
.

Contradiction.
In conclusion we have and the solution is followed.
Bai toan : ,and a point . is Cevian triangle of . are
midpoints of . cut at .
Prove that are concurrent.
Generalization:
lie on such that concur at . Then
concur at which lies on .
Solution:
Let be the intersections of and and and ,
respectively then are collinear.
.
We have then concur at .
Denote .
Applying Menelaus's theorem we obtain: .
Denote .
We will show that
(it's right from Menelaus's theorem)
Therefore . Similarly we are done.
Bai toan : Problem (zaizai-hoang):
Let a,b,c are positive number such that: . Prove that:
.
Loi giai : Nguyen Cong
Lemma. for ,
Lemma-pf) Let . then ineq is changed to
using AM-GM, we get
so ETS
Here, from Schur ineq case and AM-GM, we get
so the lemma is proved.
Let's start to prove the problem !

If then
so the condition can't be satisfied. Therefore,
Using lemma, ETS
Nhan xet: Bai toan tren con co cach giai khac la dung BDT Schur ket hop voi pqr nhung
loi giai co nhung bien doi kha dai .Khong dung voi ve dep cua bDT .
Let for which and Consider the point
such that
and the point so that Prove that
Let be the circumcircle of and . We have:
, i.e., and are
congruent.
Since, and then .
In , we have:
But so . Thus
, and we are done.
Let be trhee non-negative real numbers. Prove that