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)