Question

Cho x,y,z>0 và \(\dfrac{1}{1+x}+\dfrac{1}{1+y}+\dfrac{1}{1+z}\ge2\)

Chứng minh: xyz≤\(\dfrac{1}{8}\)

Answer 1

Ta có \(\dfrac{1}{1+x}\ge1-\dfrac{1}{1+y}+1-\dfrac{1}{1+x}=\dfrac{y}{1+y}+\dfrac{z}{1+z}\)

\(\ge2\sqrt{\dfrac{yz}{\left(y+1\right)\left(z+1\right)}}\)

Chứng minh tương tự, ta có

\(\dfrac{1}{1+y}\ge2\sqrt{\dfrac{xz}{\left(z+1\right)\left(x+1\right)}};\dfrac{1}{1+z}\ge2\sqrt{\dfrac{xy}{\left(x+1\right)\left(y+1\right)}}\)

Nhân cả 3 cua 3 BĐT cùng chiều, ta có

\(\dfrac{1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\ge\dfrac{8xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\Rightarrow xuz\le\dfrac{1}{8}\left(ĐPCM\right)\)