Given $x,y,z >0$, $1/x+1/y+1/z = 4$, prove that $ 1/(2x+y+z)+1/(x+2y+z) +1/(x+y+2z) \le 1$

Question

Given $x,y,z >0$, $1/x+1/y+1/z = 4$, prove that $$ 1/(2x+y+z)+1/(x+2y+z) +1/(x+y+2z) \le 1 .$$

Any hints or direction will be appreciated.

Answer 1

Try AM-HM inequality on each term.. The first one can be written as ${\displaystyle {1 \over {1 \over x^{-1}} + {1 \over x^{-1}} + {1 \over y^{-1}} + {1 \over z^{-1}}}}$ for example...

Answer 2

By C-S $$\sum_{cyc}\frac{1}{2x+y+z}\leq\frac{1}{(2+1+1)^2}\sum_{cyc}\left(\frac{2^2}{2x}+\frac{1^2}{y}+\frac{1^2}{z}\right)=1$$