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

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 !