Question

[Yim Yim]

cho x;y;z >0 và \(x+y+z\ge1\). Chứng minh rằng

\(\frac{x^3}{y^2}+\frac{y^3}{z^2}+\frac{z^3}{x^2}\ge1\)

Answer 1

[Đinh Đức Hùng]

Áp dụng bđt AM - GM ta có : 

\(\frac{x^3}{y^2}+x\ge2\sqrt{\frac{x^3}{y^2}.x}=\frac{2x^2}{y}\)

\(\frac{y^3}{z^2}+y\ge2\sqrt{\frac{y^3}{z^2}.y}=\frac{2y^2}{z}\)

\(\frac{z^3}{x^2}+z\ge2\sqrt{\frac{z^3}{x^2}.z}=\frac{2z^2}{x}\)

Cộng vế với vế ta được :

\(\frac{x^3}{y^2}+\frac{y^3}{z^2}+\frac{z^3}{x^2}+x+y+z\ge2\left(\frac{x^2}{y}+\frac{y^2}{z}+\frac{x^2}{z}\right)\)

Ta lại có : \(\left(\frac{x^2}{y}+\frac{y^2}{z}+\frac{x^2}{z}\right)\left(x+y+z\right)\ge\left(x+y+z\right)^2\)(bunhiacopxki)

\(\Rightarrow\frac{x^2}{y}+\frac{y^2}{z}+\frac{x^2}{z}\ge\frac{\left(x+y+z\right)^2}{x+y+z}=x+y+z\)

\(\Rightarrow\frac{x^3}{y^2}+\frac{y^3}{z^2}+\frac{z^3}{x^2}+x+y+z\ge2\left(\frac{x^2}{y}+\frac{y^2}{z}+\frac{x^2}{z}\right)\ge2\left(x+y+z\right)\)

\(\Rightarrow\frac{x^3}{y^2}+\frac{y^3}{z^2}+\frac{z^3}{x^2}\ge x+y+z\ge1\)(đpcm)