Answer
See explanations.
Work Step by Step
Step 1. Rewrite the inequality as $(a+\frac{1}{a})(b+\frac{1}{b})(c+\frac{1}{c})(d+\frac{1}{d})\geq2^4$. Considering the repetition of equivalent terms, all we need to prove is $(x+\frac{1}{x})\geq2$
Step 2. Letting $y=(x+\frac{1}{x}$, find its extrema by letting $y'=0$ to get $1-\frac{1}{x^2}=0$, which gives $x=1$ (as $x\gt0$)
Step 3. At $x=1$, $y(1)=2$. Check $y''(1)=\frac{2}{x^3}=2\gt0$; thus $y(1)=2$ is a minimum. In other words, $x+\frac{1}{x}\geq2$ over its domain.
