# An Inequality from RMM with Powers of 2

### Proof

Note that

\begin{align} 2^x+2^y+2^z+2^{x+y+z}&-2^{x+y}-2^{y+z}-2^{z+x}-1\\ &=(2^x-1)(2^y-1)(2^z-1)\\ &\gt 0, \end{align}

implying that

$2^x+2^y+2^z+2^{x+y+z}\gt 2^{x+y}+2^{y+z}+2^{z+x}+1.$

But

\begin{align} 2^{x+y}&=4^{\frac{x+y}{2}}\\ &\ge 4^{\frac{2xy}{x+y}}\\ &=\sqrt[x+y]{16^{xy}} \end{align}

The required inequality follows from the above by cycling through the pairs $(y,z)\;$ and $(z,x)\;$ then adding.

### Acknowledgment

Dan Sitaru has kindly posted the above problem from the Romanian Mathematical Magazine, with a proof by Ravi Prakash (India), at the CutTheKnotMath facebook page.

The proof appears to work for real positive $x,y,z.$

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