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

I don't think Jensen's inequality will help here, but I think first determining where equality holds will be useful. Maybe taking the logarithm or exponential of both sides will also be useful, but I want to in the end get rid of the plus signs in order to simplify it.

TL;DR: The inequality has been proven for all cases except the following five:

• $1 < a < 2$, $b < 1$, $c > 1$, $d < 1$




• $1 < a < 2$, $b < 1$, $c < 1$, $d < 1$




• $2 < a < 3$, $b < 0.207$, $c > 1$, $d < 1$




• $2 < a < 3$, $b < 0.256$, $c < 1$, $d < 1$




• $3 < a < 4$, $b < 0.129$, $c < 1$, $d < 1$

This partial answer heavily uses the results that for a real number $k>0$,

• $min x^{kx}=(sqrt[e]{e^k})^{-1}$,




• $k^{kx}>k^{kx_0}$ for $k>1$ and $x>x_0$,




• $k^{kx}<k^{kx_0}$ for $k<1$ and $x>x_0$.

As the sum $$S=a^{ab}+b^{bc}+c^{cd}+d^{da}$$ is cyclic, we need only concern $a=1$ and $a>1$ to prove the truth of the inequality. Thus there are sixteen cases that we need to consider (note that in most cases the condition $a+b+c+d=4$ is implicitly used).

$1)$ $a=b=c=d=1$

Clearly $S=1+1+1+1>pi$.

$2)$ $a=b=1$, $c>1$, $d<1$

As $c<2$, $S>1+1+1+(sqrt[e]{e})^{-1}>pi$.

$3)$ $a=b=1$, $c<1$, $d>1$

As $d<2$, $S>1+1+(sqrt[e]{e^2})^{-1}+1>pi$.

$4)$ $a=1$, $b>1$, $c>1$, $d<1$

We have $S>1+1+1+(sqrt[e]{e})^{-1}>pi$.

$5)$ $a=1$, $b>1$, $c<1$, $d>1$

As $d<2$, $S>1+1+(sqrt[e]{e^2})^{-1}+1>pi$.

$6)$ $a=1$, $b>1$, $c<1$, $d<1$

We have $S>1+1+(sqrt[e]{e})^{-1}+(sqrt[e]{e})^{-1}>pi$.

$7)$ $a=1$, $b<1$, $c>1$, $d<1$

If $bge0.6$, $cle2.4$ so $Sge1+0.6^{0.6cdot2.4}+1+(sqrt[e]{e})^{-1}>pi$. If $b<0.6$, $c>1.4$ so $S>1+(sqrt[e]{e^3})^{-1}+min{1.4^{1.4d}+d^d}>pi$.

$8)$ $a=1$, $b<1$, $c<1$, $d>1$

If $bge0.89$, $dle2.11$ so $Sge1+(sqrt[e]{e})^{-1}+(sqrt[e]{e^{2.11}})^{-1}+1>pi$. If $b<0.89$, $3>d>1.11$ so $Sge 1+(sqrt[e]{e})^{-1}+(sqrt[e]{e^3})^{-1}+1.11^{1.11}>pi$.

$9)$ $a=1$, $b<1$, $c>1$, $d>1$

As $c<2$, $S>1+(sqrt[e]{e^2})^{-1}+1+1>pi$.

$10)$ $a>1$, $b>1$, $c>1$, $d<1$

As $a<2$, $S>1+1+1+(sqrt[e]{e^2})^{-1}>pi$.

$11)$ $a>1$, $b>1$, $c<1$, $d>1$

As $d<2$, $S>1+1+(sqrt[e]{e^2})^{-1}+1>pi$.

$12)$ $a>1$, $b>1$, $c<1$, $d<1$

If $dge0.675$, $S>1+1+(sqrt[e]{e^{0.675}})^{-1}+0.675^{0.675}>pi$. If $age2$, $d<0.675$, then $S>2^2+0+0+0>pi$. If $a<2$, $d<0.675$, then $S>1+1+(sqrt[e]{e})^{-1}+(sqrt[e]{e^2})^{-1}>pi$.

$13)$ $a>1$, $b<1$, $c>1$, $d<1$

If $3>a>2$, $c<2$ so $bge0.207$, $S>2^{2cdot0.207}+(sqrt[e]{e^2})^{-1}+1+(sqrt[e]{e^3})^{-1}>pi$.

$14)$ $a>1$, $b<1$, $c<1$, $d>1$

If $dge2$, $S>0+0+0+2^2>pi$. If $d<2$, $S>1+(sqrt[e]{e})^{-1}+(sqrt[e]{e^2})^{-1}+1>pi$.

$15)$ $a>1$, $b<1$, $c>1$, $d>1$

As $c<2$, $S>1+(sqrt[e]{e^2})^{-1}+1+1>pi$.

$16)$ $a>1$, $b<1$, $c<1$, $d<1$

If $4>a>3$, $bge0.129$, $S>3^{3cdot0.129}+(sqrt[e]{e})^{-1}+(sqrt[e]{e})^{-1}+(sqrt[e]{e^4})^{-1}>pi$. If $3>a>2$, $bge0.256$, $S>2^{2cdot0.256}+(sqrt[e]{e})^{-1}+(sqrt[e]{e})^{-1}+(sqrt[e]{e^3})^{-1}>pi$.

The problem may be simplified with AM-GM inequality:

$$a^{ab}+b^{bc}+c^{cd}+d^{da}geq 4(a^{ab}b^{bc}c^{cd}d^{da})^{frac{1}{4}} geq pi $$

The equality between AM and GM is valid if and only if all of the factors are equal so we must solve:

$$a^{ab}=b^{bc}=c^{cd}=d^{da} Rightarrow (4-b-c-d)^{(4-b-c-d)b}=b^{bc}=c^{cd}=d^{d(4-b-c-d)} $$

And now we can reduce this to a system:
$$begin{cases}
(4-b-c-d)^{(4-b-c-d)}=b^{c} \
b^{b}=c^{d} \
c^{c}=d^{(4-b-c-d)} \
end{cases}
$$

Unluckly i'm an high school noob and i have only geogebra that explodes if I put one of these equations. However we must find all of the solutions of the system and verify that for each of these:

$$4(a^{ab}b^{bc}c^{cd}d^{da})^{frac{1}{4}} geq pi $$

I found only the trivial one (1,1,1) and in this case trivially $$ 4 geq pi $$