amparvardi wrote: Prove the inequality \[ \frac{a_1+ a_3}{a_1 + a_2} + \frac{a_2 + a_4}{a_2 + a_3} + \frac{a_3 + a_1}{a_3 + a_4} + \frac{a_4 + a_2}{a_4 + a_1} \geq 4, \] where $a_i > 0, i = 1, 2, 3, 4.$ Let $a,$ $b$, $c$ and $d$ are positives. Hence, $\sum_{cyc}\frac{a+c}{a+b}=(a+c)\left(\frac{1}{a+b}+\frac{1}{c+d}\right)+(b+d)\left(\frac{1}{b+c}+\frac{1}{a+d}\right)\geq$ $\geq\frac{4(a+c)}{a+b+c+d}+\frac{4(b+d)}{a+b+c+d}=4$.

Also one can notice that the inequality directly follows from Cauchy Schwarz since: \[\left(\[ \frac{a_{1}+a_{3}}{a_{1}+a_{2}}+\frac{a_{2}+a_{4}}{a_{2}+a_{3}}+\frac{a_{3}+a_{1}}{a_{3}+a_{4}}+\frac{a_{4}+a_{2}}{a_{4}+a_{1}}\right)((a_1+a_3)(a_1+a_2)+(a_2+a_4)(a_2+a_3)+(a_3+a_1)(a_3+a_4)+(a_4+a_2)(a_4+a_1))=\left(\[ \frac{a_{1}+a_{3}}{a_{1}+a_{2}}+\frac{a_{2}+a_{4}}{a_{2}+a_{3}}+\frac{a_{3}+a_{1}}{a_{3}+a_{4}}+\frac{a_{4}+a_{2}}{a_{4}+a_{1}}\right)(a_1+a_2+a_3+a_4)^2\geq[2(a_1+a_2+a_3+a_4)^2]\]$\Rightarrow$ \[ \frac{a_{1}+a_{3}}{a_{1}+a_{2}}+\frac{a_{2}+a_{4}}{a_{2}+a_{3}}+\frac{a_{3}+a_{1}}{a_{3}+a_{4}}+\frac{a_{4}+a_{2}}{a_{4}+a_{1}}\geq 4, \]