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