If a, b, c, d are positive real numbers such that $a+b+c+d=8,$find the smallest value of M such that${(a+\frac{1}{a})}^2+{(b+\frac{1}{b})}^2+{(c+\frac{1}{c})}^2+{(d+\frac{1}{d})}^2\ge M$

There are many ways to solve this problem, but probably the cleanest one is the following, based on two simple applications of the QMAM-HM inequality:
${(a+\frac{1}{a})}^2+{(b+\frac{1}{b})}^2+{(c+\frac{1}{c})}^2+{(d+\frac{1}{d})}^2\ge$

$\frac{1}{4}{(a+\frac{1}{a}+b+\frac{1}{b}+c+\frac{1}{c}+d+\frac{1}{d})}^2=\frac{1}{4}{(8+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})}^2,$

the last equality being a consequence of the hypothesis a+b+c+d=8. Hence, it suffices to prove that
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge 2$

But this is a consequence of the AM-HM inequality

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge \frac{16}{a+b+c+d}=2.$