Supremum of $a^{ab}+b^{bc}+c^{cd}+d^{da}$ with $a+b+c+d=4$

Question

Let $a,b,c,d>0$ I want to find the supremum of :

$$a^{ab}+b^{bc}+c^{cd}+d^{da}$$ With $a+b+c+d=4$

I claim that the supremum has the following form : $$a^{ab}+3$$ With $a+b=4$

In fact it remains to prove the following theorem :

Let $a,b,c,d>0$ such that $a+b+c+d=4$ and $a>b>c>d$ then we have : $$a^{ab}+b^{bc}+c^{cd}+d^{da}< a^{ab}+3$$

All of this is just an intuition and I'm really stuck to prove this...

If you have hints it will be nice .

Thanks in advance for your time .

Answer 1

Your conjecture is not true. I just generated 4 random numbers satisfying the condition $a+b+c+d=4$ and immediately found counter-examples.