This inequality why can't solve it by now (Only four variables inequality)? I asked a question at Math.SE last year and later offered a bounty for it, only johannesvalks give Part of the answer; A few months ago, I asked the author(Pham kim Hung)  in Facebook, he said that now there is no proof by hand.and use of software verification is correct,and I try it sometimes,and not succeed.Later asked a lot of people (such on AOPS 1,AOPS 2) have no proof
interesting inequality:
Let $a,b,c,d>0$, show that
$$\dfrac{1}{4}\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)\ge \sqrt[4]{\dfrac{a^4+b^4+c^4+d^4}{4}}$$
In fact,we have
$$\dfrac{1}{4}\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)\ge \underbrace{\sqrt[4]{\dfrac{a^4+b^4+c^4+d^4}{4}}\ge \sqrt{\dfrac{a^2+b^2+c^2+d^2}{4}}}_{\text{Generalized mean}}$$
Now we only prove this not stronger inequality:
$$\dfrac{1}{4}\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)\ge \sqrt{\dfrac{a^2+b^2+c^2+d^2}{4}}$$
Proof:By Holder inequality we have
$$\left(\sum_{cyc}\dfrac{a^2}{b}\right)^2(a^2b^2+b^2c^2+c^2d^2+d^2a^2)\ge (a^2+b^2+c^2+d^2)^3$$
and Note
$$a^2b^2+b^2c^2+c^2d^2+d^2a^2=(a^2+c^2)(b^2+d^2)\le\dfrac{(a^2+b^2+c^2+d^2)^2}{4}$$
Proof 2:(I hope following  methods(creat is Mine) will usefull to solve my OP inequality,So I post it):
\begin{align*}&\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)^2-4(a^2+b^2+c^2+d^2)\\
&=\sum_{cyc}\dfrac{3a^4b^2d+5a^4c^3+24a^3cd^3+3a^2b^3c^2+10ab^3d^3+15bcd^5-60a^2bcd^3}{15a^2bcd}\\
&\ge 0
\end{align*}
NoW I use computer 
$$\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)^4-64(a^2+b^2+c^2+d^2)=\dfrac{a^{12}c^4d^4+4a^{10}b^3c^3d^4+4a^{10}bc^6d^3+4a^9bc^4d^6
+6a^8b^6c^2d^4+12a^8b^4c^5d^3+64a^8b^4c^4d^4+6a^8b^2c^8d^2+12a^7b^4c^3d^6+12a^7b^2c^6d^5+4a^6b^9cd^4+\cdots+4ab^4c^6d^9+b^4c^4d^{12}}{a^4b^4c^4d^4}$$

 A: Here is a partial solution that reduces the problem to a (hopefully) simpler one.
The inequality is homogeneous, so we may assume that the RHS equals one.
Let
$$
S=\left\{x\in\mathbb R^4;\frac14\sum_ix_i^4=1,x_i>0\right\}
$$
and
$$
f(a,b,c,d)=\dfrac{1}{4}\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right).
$$
Clearly $f$ and $S$ are smooth, $S$ is bounded and $f(x)\to\infty$ as $x$ approaches any boundary point, so it suffices to show that the inequality is satisfied at points given by Lagrange's multiplier theorem.
We get the conditions $(a,b,c,d)\in S$ and
\begin{eqnarray}
&&
\frac14
\left(
2\frac ab-\left(\frac da\right)^2,
2\frac bc-\left(\frac ab\right)^2,
2\frac cd-\left(\frac bc\right)^2,
2\frac da-\left(\frac cd\right)^2
\right)
\\&=&
\lambda
(a^3,b^3,c^3,d^3)
\end{eqnarray}
for some $\lambda\in\mathbb R$.
Clearly we need to have $\lambda>0$.
Taking the dot product with the vector $(a,b,c,d)$ we obtain
$$
f(a,b,c,d)
=
4\lambda
$$
so it suffices to show that $\lambda\geq\frac14$.
If this were not the case, we would have
\begin{eqnarray}
&&
\left(
2\frac ab-\left(\frac da\right)^2,
2\frac bc-\left(\frac ab\right)^2,
2\frac cd-\left(\frac bc\right)^2,
2\frac da-\left(\frac cd\right)^2
\right)
\\&>&
(a^3,b^3,c^3,d^3),
\end{eqnarray}
meaning inequality for each component.
It would now suffice to show that this is impossible if $(a,b,c,d)\in S$.
A: As to why the question is hard - one could reformulate it as $$\frac{x+y+z+t}{4}\stackrel{?}{\ge} \sqrt[4]{\frac{(x^8 y^4 z^2 t)^{4/15}+(y^8 z^4 t^2 x)^{4/15}+(z^8 t^4 x^2 y)^{4/15}+(t^8 x^4 y^2 z)^{4/15}}{4}}$$
where $x=a^2/b$, $y=b^2/c$, $z=c^2/d$ and $t=d^2/a$. Then there is a tug of war between
$$\frac{x+y+z+t}{4}\le \sqrt[4]{\frac{x^4+y^4+z^4+t^4}{4}} $$
and
$$\frac{x+y+z+t}{4}\ge \frac{(x^8 y^4 z^2 t)^{1/15}+(y^8 z^4 t^2 x)^{1/15}+(z^8 t^4 x^2 y)^{1/15}+(t^8 x^4 y^2 z)^{1/15}}{4}$$
with the latter apparently winning.
