# Prove that $\sum_{cyc}^{}\sqrt{a^{2}+1}\leq\sqrt{2}\sum_{cyc}^{} a$

$a$,$b$,$c$,$d>0$
$abc+bcd+cda+dab=a+b+c+d$

Prove the following inequality

$\sqrt{a^{2}+1}+\sqrt{b^{2}+1}+\sqrt{c^{2}+1}+\sqrt{d^{2}+1}\leq\sqrt{2}(a+b+c+d)$

• Probably you can use $a^2+1=\frac{a^2(a+b+c+d)+abc+bcd+cda+dab}{a+b+c+d}=\frac{(a+b)(a+c)(a+d)}{a+b+c+d}$. – Bart Michels Mar 14 '16 at 10:01
• – Martin Sleziak May 27 '18 at 18:59

Easy to see that $\sum\limits_{cyc}\frac{a^2+1}{a+b}=a+b+c+d$.
Hence, by C-S $a+b+c+d=\sum\limits_{cyc}\frac{a^2+1}{a+b}\geq\frac{\left(\sum\limits_{cyc}\sqrt{a^2+1}\right)^2}{2(a+b+c+d)}$ and we are done!