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 $b\ge0.6$, $c\le2.4$ so $S\ge1+0.6^{0.6\cdot2.4}+1+(\sqrt[e]{e})^{-1}>\pi$. 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 $b\ge0.89$, $d\le2.11$ so $S\ge1+(\sqrt[e]{e})^{-1}+(\sqrt[e]{e^{2.11}})^{-1}+1>\pi$. If $b<0.89$, $3>d>1.11$ so $S\ge 1+(\sqrt[e]{e})^{-1}+(\sqrt[e]{e^3})^{-1}+1.11^{1.11}>\pi$.
$9)$ $a=1$, $b<1$, $c>1$, $d>1$
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 $d\ge0.675$, $S>1+1+(\sqrt[e]{e^{0.675}})^{-1}+0.675^{0.675}>\pi$. If $a\ge2$, $d<0.675$, 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 $b\ge0.207$, $S>2^{2\cdot0.207}+(\sqrt[e]{e^2})^{-1}+1+(\sqrt[e]{e^3})^{-1}>\pi$.
$14)$ $a>1$, $b<1$, $c<1$, $d>1$
If $d\ge2$, $S>0+0+0+2^2>\pi$. 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 $4>a>3$, $b\ge0.129$, $S>3^{3\cdot0.129}+(\sqrt[e]{e})^{-1}+(\sqrt[e]{e})^{-1}+(\sqrt[e]{e^4})^{-1}>\pi$. If $3>a>2$, $b\ge0.256$, $S>2^{2\cdot0.256}+(\sqrt[e]{e})^{-1}+(\sqrt[e]{e})^{-1}+(\sqrt[e]{e^3})^{-1}>\pi$.
Best Answer
We can rewrite $x^2+y^2+1 \geq xy+x+y \ $ as
$2x^2+2y^2+2 - 2xy - 2x - 2y \geq 0$
or as $(x-y)^2 + (x-1)^2 + (y-1)^2 \geq 0$
which holds for all $x, y \in \mathbb{R}$
Or start from $(x-y)^2 + (x-1)^2 + (y-1)^2 \geq 0$ and expand to show that $x^2+y^2+1 \geq xy+x+y \ $.