Školjka

\begin{align*} \frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b} \ge 0\\ \Leftrightarrow \frac{a-b+b+c}{b+c}+\frac{b-c+c+d}{c+d}+\frac{c-d+d+a}{d+a}+\frac{d-a+a+b}{a+b} \ge 4\\ \Leftrightarrow \frac{a+c}{b+c}+\frac{b+d}{c+d}+\frac{c+a}{d+a}+\frac{d+b}{a+b} \ge 4\\ \Leftrightarrow (a+c)(\frac{1}{b+c}+\frac{1}{a+d})+(b+d)(\frac{1}{c+d}+\frac{1}{a+b}) \ge 4\\ \end{align*} Po Engel formi CSB nejednakosti imamo \begin{align*} \frac{1}{b+c}+\frac{1}{a+d} \ge \frac{(1+1)^2}{a+b+c+d}=\frac{4}{a+b+c+d}, \ \ \frac{1}{c+d}+\frac{1}{a+b} \ge \frac{(1+1)^2}{a+b+c+d}=\frac{4}{a+b+c+d}\\ \end{align*} Sad uvrštavanjem dobijemo:\begin{align*} (a+c)(\frac{1}{b+c}+\frac{1}{a+d})+(b+d)(\frac{1}{c+d}+\frac{1}{a+b}) \ge (\frac{4}{a+b+c+d})(a+b+c+d)=4 \end{align*}