Question

cho x,y,z là các số thực dương 

CMR: P= \(\frac{3\left(x^3+y^3+z^3\right)}{4\left(xy+yz+zx\right)}+\frac{1}{\left(x+y+z\right)^2}\ge\frac{3}{4}\)

Answer 1

\(P=\frac{3\left(x^3+y^3+z^3\right)}{4\left(xy+yz+zx\right)}+\frac{1}{\left(x+y+z\right)^2}\ge\frac{\left(x+y+z\right)\left(xy+yz+zx\right)}{4\left(xy+yz+zx\right)}+\frac{1}{\left(x+y+z\right)^2}\)

\(=\frac{x+y+z}{4}+\frac{1}{\left(x+y+z\right)^2}\)

Đặt \(x+y+z=a\) thì cần chứng minh

\(\frac{a}{4}+\frac{1}{a^2}\ge\frac{3}{4}\)

\(\Leftrightarrow\left(a-2\right)^2\left(a+1\right)\ge0\)(đúng)