Prove the inequality $(1+2^{x-1}+2^y)(1+2^{y-1}+2^z)(1+2^{z-1}+2^x)\leq 2^{x^2+y^2+z^2+3}$

Question

question

Let $x,y,z$ natural nonzero numbers. Prove that:

$(1+2^{x-1}+2^y)(1+2^{y-1}+2^z)(1+2^{z-1}+2^x)\leq 2^{x^2+y^2+z^2+3}$

my idea

I don't know why but the first thing I thought of was to break down the parentheses, but I don't think this is a really good idea.

I know this type of inequality is proved using the inequality of means or CBS. To be honest, this inequality reminds me of AM-GM.

The thing is I never know where to start from. Hope one of you can help me! Thank you!

edit

Without loss of generality, I let $x\leq y\leq z$ (I don't think this is going to help us that much, but still)

by AM-GM

$\sqrt[\leftroot{10} \uproot{5} 3]{(1+2^{x-1}+2^y)(1+2^{y-1}+2^z)(1+2^{z-1}+2^x)} \leq \frac{3+2^{x-1}(1+2)+2^{y-1}(1+2)+2^{z-1}(1+2)}{3}$

$\sqrt[\leftroot{10} \uproot{5} 3]{(1+2^{x-1}+2^y)(1+2^{y-1}+2^z)(1+2^{z-1}+2^x)} \leq 1+2^{x-1}+2^{y-1}+2^{z-1}$

$(1+2^{x-1}+2^y)(1+2^{y-1}+2^z)(1+2^{z-1}+2^x) \leq {(1+2^{x-1}+2^{y-1}+2^{z-1})}^3$

Answer 1

I think Calvin Lin's idea might be worth an answer here.

First replace with $2^{x-1} + 1\le 2^x$, then the product becomes 8 terms. Without loss of generality, let $x\ge y\ge z \ge 1$.

The largest term is $2^{2x + y}$, if $x=1$, then $x=y=z=1$, the inequality is true. So just focus on $x\ge 2$ (since all are integers), then

$$2^{2x + y} \le 2^{x^2 + y^2 + z^2}$$

which means the product is $\le 8 \cdot 2^{x^2 + y^2 + z^2}$.