# Difficult inequality with $\pi$

My question is about an inequality ,originally I wanted to prove this :

If $a,b,c,d >0$, and $a+b+c+d=4$, prove that $$a^{ab}+b^{bc}+c^{cd}+d^{da} \geq \pi.$$ from here

My approach is to use this two inequality with the same conditions :

If $a,b,c,d >0$, and $a+b+c+d=4$,

$\frac{\pi}{2}(ac)^{abcd}\leq a^{ab}+d^{da}$

And

$\frac{\pi}{2}(bd)^{abcd}\leq b^{bc}+c^{cd}$

But I have no idea to prove this two last inequality...Thanks!

• Probably it will be adequate the tag "number theory" here. – Masacroso Mar 2 '17 at 12:48
• We can make something stronger: $a^{ab}+b^{bc}+c^{cd}+d^{da}>3.16$ – Michael Rozenberg Mar 2 '17 at 12:59
• @ Michael Rozenberg You surely direct the OP to $\sqrt{10}$... – Jean Marie Mar 2 '17 at 13:09
• @JeanMarie For $\sqrt{10}$ the starting inequality is wrong. – Michael Rozenberg Mar 2 '17 at 17:17
• @Michael Rozenberg, could you post the proof of the original inequality? – Ivan Mar 6 '17 at 4:56