Question

Cho x,y,z là các số dương \(\le1\). Chứng minh rằng : \(\frac{x}{1+y+xz}+\frac{y}{1+z+xy}+\frac{z}{1+x+yz}\le\frac{3}{x+y+z}\)

GIÚP MÌNH NHA!...

Answer 1

Áp dụng BĐT Cauchy-Schwarz ta có:

\(\frac{x}{1+y+xz}=\frac{x\left(x^2+y+\frac{z}{x}\right)}{\left(1+y+xz\right)\left(x^2+y+\frac{z}{x}\right)}\le\frac{x^3+xy+z}{\left(x+y+z\right)^2}\)

\(\le\frac{x+y+z}{\left(x+y+z\right)}=\frac{1}{x+y+z}\)

Tương tự ta cũng có: \(\frac{y}{1+z+xy}\le\frac{1}{x+y+z};\frac{z}{1+x+yz}\le\frac{1}{x+y+z}\)

Cộng theo vế ta có: \(\frac{x}{1+y+xz}+\frac{y}{1+z+xy}+\frac{z}{1+x+yz}\le\frac{1+1+1}{x+y+z}=\frac{3}{x+y+z}\)