\(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=\left(\frac{x}{y+z}+1\right)+\left(\frac{y}{x+z}+1\right)+\left(\frac{z}{x+y}+1\right)-3\)
\(=\frac{x+y+z}{y+z}+\frac{x+y+z}{x+z}+\frac{x+y+z}{x+y}-3=\left(x+y+z\right).\left(\frac{1}{y+z}+\frac{1}{x+z}+\frac{1}{x+y}\right)-3\)
\(=\frac{1}{2}\left[\left(x+y\right)+\left(y+z\right)+\left(x+z\right)\right]\left(\frac{1}{y+z}+\frac{1}{x+z}+\frac{1}{x+y}\right)-3\ge\frac{9}{2}-3=\frac{3}{2}\left(đpcm\right)\)
\(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}=\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)+\left(\frac{x}{z}+\frac{z}{x}\right)\)
Áp dụng Cô-Si cho các số không âm:
\(\frac{x}{y}+\frac{y}{x}\ge2\sqrt{\frac{x}{y}.\frac{y}{x}}=2;\frac{y}{z}+\frac{z}{y}\ge2\sqrt{\frac{y}{z}.\frac{z}{y}}=2;\frac{x}{z}+\frac{z}{x}\ge2\sqrt{\frac{x}{z}.\frac{z}{x}}=2\)
Cộng theo vế các bất đẳng thức ta được: \(\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)+\left(\frac{x}{z}+\frac{z}{x}\right)\ge2+2+2=6\)
Xem lại đề...............
\(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=\left(\frac{x}{y+z}+1\right)+\left(\frac{y}{x+z}+1\right)+\left(\frac{z}{x+y}+1\right)-3\)
\(=\frac{x+y+z}{y+z}+\frac{x+y+z}{x+z}+\frac{x+y+z}{x+y}-3=\left(x+y+z\right).\left(\frac{1}{y+z}+\frac{1}{x+z}+\frac{1}{x+y}\right)-3\)
Tới đây bn thử C/m \(\left(x+y+z\right).\left(\frac{1}{y+z}+\frac{1}{x+z}+\frac{1}{x+y}\right)\ge\frac{9}{2}\) xem,trừ 3 sẽ đc đpcm.................
Bạn chỉ cần áp dụng bdt (a+b+c)(1/a + 1/b + 1/c) >= 9 . cm rat nhanh luon -_-
Ap dung bdt tren ta dc : ( y+ c/x + x+ c/y + x+y/c )(x/y+c + y/x+c + c/x+y )>= 9 tuong duong : ( Y/X + X/Y + Y/ C + C/Y + X/C+ C /X)(X/Y+C + Y/X+C + C/X+Y) >=9 MA X/Y + Y/X >=2 , X/C + C /X >=2 , Y/C + C/Y>=2 VOI MOI X,Y,C>0 DO DO 6(X/Y+C + Y/X+C + C/X+Y ) >=9 TUONG DUONG X/Y+C + Y/X+C + C/X+Y >= 3/2 Thong cam nha cai ban phim k bấm đc chữ ''dét'' -_-
