# Showing $64\le\left(1+\frac1x\right)\left(1+\frac1y\right)\left(1+\frac1z\right)$

Showing $64\le\left(1+\frac1x\right)\left(1+\frac1y\right)\left(1+\frac1z\right)$ subject to $x,y,z>0$ and $x+y+z=1$

This inequality is equivalent to;

$\left(\frac{x+1}{4}\cdot\frac1x\right)\left(\frac{y+1}{4}\cdot\frac1y\right)\left(\frac{z+1}{4}\cdot\frac1z\right)\ge1$ taking logarithm of both sides we have to show that LHS is greater or equal to $0$, but by convexity of $\ln\left(\frac{x+1}{4}\right)-\ln(x)$, i.e. $\frac{d^2}{dx^2}\left[\ln\left(\frac{x+1}{4}\right)-\ln(x)\right]= \frac{1}{x^2}-\frac{1}{(x+1)^2}>0$ we have

$\sum\limits_{cyc}\ln\left(\frac{x+1}{4}\right)-\ln(x)\ge3\ln\left(\sum\limits_{cyc}\frac{x+1}{4}\right)-3\ln\left(\sum\limits_{cyc}x\right)=3\ln(1)-3\ln(1)=0$

Is this OK, or is there a much simpler way ? (I did the same as in example $4.1.2.$ here)

By AM-GM we have: $$\frac{1}{3}=\frac{1}{3}(x+y+z)\geq\sqrt{xyz}\implies\frac{1}{27}\geq xyz\implies\frac{1}{27xyz}\geq 1.$$ Using AM-GM again, with 4 terms this time, we obtain $$1+\frac{1}{x}=1+\frac{1}{3x}+\frac{1}{3x}+\frac{1}{3x}\geq 4\sqrt{\frac{1}{27x^3}}.$$ Do that for the other 2 terms and multiply. We get $$\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right)\geq 64\sqrt{\left(\frac{1}{27xyz}\right)^3}\geq 64.$$ Equalities are iff $x=y=z=\frac{1}{3}$.
By Holder and AM-GM $\prod\limits_{cyc}\left(1+\frac{1}{x}\right)\geq\left(1+\frac{1}{\sqrt{xyz}}\right)^3\geq\left(1+\frac{1}{\frac{1}{3}}\right)^3=64$