Prove that $a^{ab}+b^{bc}+c^{cd}+d^{da} \geq \pi$ 
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.
 A: Warning : It's a partial proof but I think it's interesting .
We have the following theorem :

Let $a,b,c,d>0$ such that $abcd\geq e^{-2}$ then we have :
  $$a^{ab}+b^{bc}+c^{cd}+d^{da}\geq 4(e^{-0.5})^{e^{-1}}$$

Proof :
We begin to apply AM-GM to get :
$$a^{ab}+b^{bc}+c^{cd}+d^{da}\geq 4(a^{ab}b^{bc}c^{cd}d^{da})^{0.25}$$
Hence we want to prove :
$$4(a^{ab}b^{bc}c^{cd}d^{da})^{0.25}\geq 4(e^{-0.5})^{e^{-1}}$$
Now we take the logarithm :
$$0.25\sum_{cyc}ab\ln(a)\geq -0.5e^{-1}$$
But $x\ln(x)$ is convex so we get :
$$0.25\sum_{cyc}ab\ln(a)\geq 0.25(ab+bc+cd+da)\ln(\frac{ab+bc+cd+da}{a+b+c+d})$$
Hence we want to prove :
$$0.25(ab+bc+cd+da)\ln(\frac{ab+bc+cd+da}{a+b+c+d})\geq -0.5e^{-1}$$
Wich is true because :
$$\frac{ab+bc+cd+da}{a+b+c+d}\geq e^{-0.5}$$
And 
$$a+b+c+d\geq 4e^{-1}$$
Now remains to add the condition $a+b+c+d=4$ to get :

Let $a,b,c,d>0$ such that $abcd\geq e^{-2}$ and $a+b+c+d=4$ then we have :
  $$a^{ab}+b^{bc}+c^{cd}+d^{da}> 4(e^{-0.5})^{e^{-1}}$$

An other idea will be to extend the initial theorem (using the same technique) and see what happend .
Finally we conclude with the fact that it completes partialy the result of TheSimpliFire .
A: We can easily prove a weaker inequality as follows. It is known that if $x>0$ then $x^x\ge \left(\tfrac 1e\right)^{1/e}=E$. Since the original inequality is cyclic, witout loss of generality we can assume that $d\ge 1$. Then 
$$a^{ab}+b^{bc}+c^{cd}+d^{da}\ge E^b+E^c+E^d+1^a\ge 3E^{(b+c+d)/3}+1\ge 
3E^{4/3}+1>2.8369.$$
A: With computer, we can use the branch and bound strategy to prove the inequality.
Proof: The inequality is written as
$$F(a, b, c, d) = \mathrm{e}^{ba\ln a} + \mathrm{e}^{cb\ln b} + \mathrm{e}^{dc\ln c} + \mathrm{e}^{a d\ln d} > \pi.$$
We first introduce some auxiliary results (Facts 1 through 2). The proofs are given later.
Fact 1: Given four real numbers $0 \le A < B, \ 0 \le C < D$ with $B-A \le \frac{1}{8}, \ D-C\le \frac{1}{8}$. Let $x\in [A, B]$ and $y\in [C, D]$. For convenience, we set $0\ln 0 = 0$. We have
i) If $\frac{1}{\mathrm{e}} \in [A, B]$, then $yx\ln x \ge -D \mathrm{e}^{-1}$;
ii) If $\frac{1}{\mathrm{e}} \notin [A, B]$ and $ A < 1$, then $yx\ln x \ge D \min(A\ln A, B\ln B)$;
iii) If $A \ge 1$, then $yx\ln x \ge C A \ln A$.
Fact 2: If $a\in [0, 1], \ b \in [0, 2], \ c \in [0, 1], \ d\ge 0$ and $a+b+c+d=4$, then $F(a,b,c,d) > \pi$.
Now let us proceed. WLOG, assume that $d= \max(a,b,c,d)$. Clearly $b \le 2$.
If $a > 1$, we have
$$\mathrm{e}^{ba\ln a} + \mathrm{e}^{cb\ln b} + \mathrm{e}^{dc\ln c} + \mathrm{e}^{a d\ln d} \ge 2\mathrm{e}^{\frac{1}{2}(cb\ln b + dc\ln c)} + 2
\ge 2\mathrm{e}^{\frac{1}{2}(-\frac{3}{\mathrm{e}})} + 2 > \pi$$
since $x\ln x \ge -\frac{1}{\mathrm{e}}$ for all $x > 0$, and $x\mapsto \mathrm{e}^x$ is convex.
If $c > 1$, we have
$$\mathrm{e}^{ba\ln a} + \mathrm{e}^{cb\ln b} + \mathrm{e}^{dc\ln c} + \mathrm{e}^{a d\ln d} \ge 2\mathrm{e}^{\frac{1}{2}(ba\ln a + cb\ln b)} + 2
\ge 2\mathrm{e}^{\frac{1}{2}(-\frac{b+c}{\mathrm{e}})} + 2
\ge 2\mathrm{e}^{\frac{1}{2}(-\frac{8/3}{\mathrm{e}})} + 2 > \pi.$$
If $a \le 1$ and $c\le 1$, from Fact 2, the desired result follows.
We are done.
$\phantom{2}$
Proof of Fact 1: Let $f(x) = x\ln x$. Then $f(x)$ is strictly decreasing on $[0, \frac{1}{\mathrm{e}})$ and strictly increasing on $(\frac{1}{\mathrm{e}}, \infty)$.
Also $f(x)$ achieves its global minimum at $x = \frac{1}{\mathrm{e}}$. It is easy to prove the desired results.
Proof of Fact 2: We use the branch and bound strategy.
Let
$$\Omega = \{(a,b,c,d): \ a\in [0, 1], \ b \in [0, 2], \ c \in [0, 1], \ d\ge 0, \ a+b+c+d=4\}.$$
For $i=0, 1, \cdots, 127; \ j=0, 1, \cdots, 255; \ k = 0, 1, \cdots, 127$, let
\begin{align}
S(i,j,k) &= \Big\{(a,b,c,d): \
a \in \Big[\frac{i}{128}, \ \frac{i+1}{128}\Big],
\quad b \in \Big[\frac{j}{128}, \ \frac{j+1}{128}\Big], \\
&\qquad\quad c \in \Big[\frac{k}{128}, \ \frac{k+1}{128}\Big], \quad
d \in \Big[4 - \frac{3 + i + j + k}{128}, \ 4 - \frac{i + j + k}{128}\Big]\Big\}.
\end{align}
We have $\Omega \subset \bigcup_{i,j,k} S(i,j,k)$.
From Fact 1, we evaluate a lower bound $L(i,j,k)$ of $F(a,b,c,d)$ on $S(i,j,k)$, i.e.,
$$F(a,b,c,d) \ge L(i,j,k), \quad \forall (a,b,c,d) \in S(i,j,k).$$
Thus, we have
$$ F(a,b,c,d) \ge \min_{i,j,k} L(i,j,k), \quad \forall (a,b,c,d)\in \Omega.$$
With computer, we obtain $\min_{i,j,k} L(i,j,k) > \pi$. The desired result follows.
This completes the proof of Fact 2.
