TITLE: This inequality why can't solve it by now (Only four variables inequality)?
QUESTION [24 upvotes]: 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 
$$\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)^4-64(a^2+b^2+c^2+d^2)=\dfrac{a^{12}c^4d^4+4a^{10}b^3c^3d^4+4a^{10}bc^6d^3+4a^9bc^4d^6
+6a^8b^6c^2d^4+12a^8b^4c^5d^3+64a^8b^4c^4d^4+6a^8b^2c^8d^2+12a^7b^4c^3d^6+12a^7b^2c^6d^5+4a^6b^9cd^4+\cdots+4ab^4c^6d^9+b^4c^4d^{12}}{a^4b^4c^4d^4}$$

REPLY [8 votes]: $\newcommand{\res}{\operatorname{res}}$
Here is a complete proof of the inequality in question.
By symmetry, without loss of generality (wlog) $d$ is the maximum of $\{a,b,c,d\}$. By homogeneity, wlog $d=1$. So, the inequality reduces to
\begin{equation*}
    r(a,b,c)
    :=\frac{\frac{a^2}{b}+\frac{b^2}{c}+c^2+\frac{1}{a}}{\sqrt[4]{a^4+b^4+c^4+1}}\ge2\sqrt2 \tag{1}
\end{equation*}
for $a,b,c$ in $(0,1]$. Consider the following polynomials in $a,b,c$:
\begin{align*}
    D_ar(a,b,c)&:=\frac{\partial r(a,b,c)}{\partial a}\, 
    a^2 b c \left(a^4+b^4+c^4+1\right)^{5/4} \\
    D_br(a,b,c)&:=\frac{\partial r(a,b,c)}{\partial b}\, 
    a b^2 c \left(a^4+b^4+c^4+1\right)^{5/4} \\
    D_cr(a,b,c)&:=\frac{\partial r(a,b,c)}{\partial c}\, 
    a b c^2 \left(a^4+b^4+c^4+1\right)^{5/4}, \\
    p_1(a,b)&:=a^{16} b^4+2 a^{16} b^2+4 a^{13} b^3+3 a^{12} b^8+4 a^{12} b^6 \\ 
&   +4 a^{12} b^4+4 a^{12} b^2+4 a^{10} b^7+2
   a^{10} b^4+6 a^{10} b^2 \\ 
   &+8 a^9 b^7+4 a^9 b^5+8 a^9 b^3+3 a^8 b^{12}+2 a^8 b^{10} \\ 
   &+6 a^8 b^8+4 a^8 b^6+3
   a^8 b^4+2 a^8 b^2+4 a^7 b^8 \\ 
   &+4 a^7 b^3+4 a^6 b^{11}+4 a^6 b^8+4 a^6 b^7+6 a^6 b^6 \\ 
   &+4 a^6 b^4+6 a^6 b^2+4
   a^5 b^{11}+8 a^5 b^7+4 a^5 b^3 \\ 
   &+a^4 b^{16}+4 a^4 b^{12}+3 a^4 b^8+2 a^4 b^4+4 a^3 b^{12} \\ 
   &+4 a^3 b^8+4 a^3
   b^7+4 a^3 b^3+2 a^2 b^{12}+4 a^2 b^8 \\ 
   &+2 a^2 b^4+4 a^{13} b+4 a^9 b+a^{16}+a^{12}+b^8+b^4, 
\end{align*}
which latter is manifestly $>0$.
Then
\begin{equation*}
    p_{11}(a,b):=\frac1{a b^3 p_1(a,b)}\,\res_c(D_ar(a,b,c),D_br(a,b,c))
\end{equation*}
is a polynomial in $a,b$ of respective degrees $35,25$, where $\res_c(D_ar(a,b,c),D_br(a,b,c))$ is the resultant with respect to $c$ of the polynomials $D_ar(a,b,c)$ and $D_br(a,b,c)$ in $a,b,c$. Similarly,
\begin{equation*}
    p_{21}(a,b):=\frac1{a^4 b^3 p_1(a,b)}\,\res_c(D_br(a,b,c),D_cr(a,b,c))
\end{equation*}
is a polynomial in $a,b$ of respective degrees $38,39$.
The key observation is that all the roots $(a,b)\in(0,1)\times(0,1)$ of the polynomials $p_{11}$ and $p_{21}$ satisfy the inequalities $b<a^2$ and $b\ge a^2$, respectively. (This was proved using the Mathematica command Reduce, which took about 6.5 sec for $p_{11}$ and 33 sec for $p_{21}$.) Thus, $p_{11}$ and $p_{21}$ have no common root $(a,b)\in(0,1)\times(0,1)$, and hence $r$ has no critical points in the cube $C:=(0,1)^3$.
It remains to consider the behavior of $r$ at/near the boundary of the cube $C$.
If $a\downarrow0$, then, by (1), $r(a,b,c)\to\infty$ uniformly in $(b,c)\in(0,1)\times(0,1)$.
If $b\downarrow0$, then, by (1), $r(a,b,c)\to\infty$ uniformly in $(a,c)\in(0,1)\times(0,1)$ unless $a\downarrow0$, which reduces the consideration to the previous paragraph.
If $c\downarrow0$, then, by (1), $r(a,b,c)\to\infty$ uniformly in $(a,b)\in(0,1)\times(0,1)$ unless $b\downarrow0$, which reduces the consideration to the previous paragraph.
It remains to consider $r(a,b,c)$ when at least one of the variables $a,b,c$ takes value $1$. In such cases, using again the Mathematica command Reduce, in splits of a second we get (1). $\Box$

Here is an image of a Mathematica notebook with details of calculations (click on the image to enlarge it):