Prove that for all positive $x,y,z$, $\small\left(2e^x+\dfrac{2}{e^x}\right)\left(2e^y+\dfrac{2}{e^y}\right)\left(2e^z+\dfrac{2}{e^z}\right) \geq 64$ Prove that for all positive $x,y,z$, $\left(2e^x+\dfrac{2}{e^x}\right)\left(2e^y+\dfrac{2}{e^y}\right)\left(2e^z+\dfrac{2}{e^z}\right) \geq 64$
I dont have that much experience with inequalities but I know I can rewrite $64 $ as $4^3$
So here is my approach, if I can maybe prove that $\left(2e^x+\dfrac{2}{e^x}\right) \geq4$ as well for $y,z$ and then maybe I can prove it is greater than 64. Any ideas?

Let $u = e^x$.  Then if $x > 0$, then $u > e^0 = 1$, and consider $$u + u^{-1} - 2 = u - 2u^{-1/2}u^{1/2} + u^{-1} = (u^{1/2} - u^{-1/2})^2 \ge 0.$$  The first equality is because $u^{-1/2}u^{1/2} = 1$, and the inequality is due to the fact that the square of a real number is never negative.  Consequently, $$u+u^{-1} \ge 2,$$ from which it follows that $$2\left(e^x + e^{-x}\right) \ge 4,$$ and the result is proven.
