Question

Cho các số thực dương \(x,y,z\)thỏa mãn \(x^2+y^2+z^2=3\)

Chứng minh rằng :\(\frac{x}{3-yz}+\frac{y}{3-xz}+\frac{z}{3-xy}\le\frac{3}{2}\)

Answer 1

\(\frac{x}{3-yz}+\frac{y}{3-zx}+\frac{z}{3-xy}\le\frac{x}{3-\frac{y^2+z^2}{2}}+\frac{y}{3-\frac{z^2+x^2}{2}}+\frac{z}{3-\frac{x^2+y^2}{2}}\)

\(=\frac{2x}{3+x^2}+\frac{2y}{3+y^2}+\frac{2z}{3+z^2}\le\frac{2x}{4\sqrt[4]{x^2}}+\frac{2y}{4\sqrt[4]{y^2}}+\frac{2z}{4\sqrt[4]{z^2}}\)

\(=\frac{\sqrt{x}}{2}+\frac{\sqrt{y}}{2}+\frac{\sqrt{z}}{2}\le\frac{x^2+3}{8}+\frac{y^2+3}{8}+\frac{z^2+3}{8}\)

\(=\frac{3}{8}+\frac{9}{8}=\frac{3}{2}\)

Answer 2

cách khác: cũng đến chỗ <= sigma 2x/3+x^2 

<= 2x/2(x+1) (do x^2+3=x^2+1+2>=2x+2) <= sigma x/x+1 = 3- sigma (1/x+1) 

sigma 1/x+1 >= 9/x+y+z+3 dễ rồi