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Let $a,b,c,d>0$ show that $$\color{blue}{\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)^2\ge 4(a^2+b^2+c^2+d^2)+\dfrac{8}{3}[(a-b)^2+(a-c)^2+(a-d)^2+(b-c)^2+(b-d)^2+(c-d)^2]}$$

By Now, I claim prove that $$\color{red}{\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)^2\ge 4(a^2+b^2+c^2+d^2)}\tag{1}$$ Proof:By $\color{red} {Holder}$ 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 $$a^2b^2+b^2c^2+c^2d^2+d^2a^2=(a^2+c^2)(b^2+d^2)\le\dfrac{(a^2+b^2+c^2+d^2)^2}{4}$$ so $(1)$ prove by Done !

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    $\begingroup$ It seems that $$\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a}\right)^2\geq13(a^2+b^2+c^2+d^2)-6(ab+ac+ad+bc+bd+cd)$$ is also true $\endgroup$ – Michael Rozenberg Jun 18 '16 at 9:09
  • $\begingroup$ To Michael Rozenberg: Try $a=4, b=3, c=2, d=2$? $\endgroup$ – River Li Jul 22 '19 at 4:32
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Update

A possible way is as follows. We use the Buffalo Way. WLOG, assume that $d = \min(a,b,c,d)$. Let $c=d+s, \ b = d+t, \ a = d+r; \ s, t, r \ge 0$. We have $\mathrm{LHS} - \mathrm{RHS} = a_8d^8 + a_7d^7 + \cdots + a_1d + a_0$. To prove (or disprove) that $a_i\ge 0$ for $i=0, 1, 2, \cdots, 8$. I verified all $a_i$'s by SOS (Sum of Squares) numerical software.

Previously written

This is not an answer. I want to discuss the relationship between two inequalities. Refer to this link How to prove $\frac{1}{4}(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a})\ge \sqrt[4]{\frac{a^4+b^4+c^4+d^4}{4}}$

Let $a, b, c, d > 0$. Prove that $$\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)^2\ge 4(a^2+b^2+c^2+d^2)+\dfrac{8}{3}[(a-b)^2+(a-c)^2+(a-d)^2+(b-c)^2+(b-d)^2+(c-d)^2], \qquad (1)$$ $$\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a} \ge 4\sqrt[4]{\frac{a^4+b^4+c^4+d^4}{4}}.\qquad\qquad\qquad (2)$$

(1) is stronger than (2) since \begin{align} &\Big(4(a^2+b^2+c^2+d^2)+\dfrac{8}{3}[(a-b)^2+(a-c)^2+(a-d)^2+(b-c)^2+(b-d)^2+(c-d)^2]\Big)^2\\ &\qquad- 64(a^4+b^4+c^4+d^4)\\ =\ & \frac{16}{9} z^TQz\\ \ge\ & 0 \end{align} where $$z = \left(\begin{array}{c} { {a}}^2\\ {a} {b}\\ { {b}}^2\\ {a} {c}\\ {b} {c}\\ { {c}}^2\\ {a} {d}\\ {b} {d}\\ {c} {d}\\ { {d}}^2 \end{array}\right), \quad Q = \left(\begin{array}{cccccccccc} 45 & -36 & -4 & -36 & 25 & -4 & -36 & 25 & 25 & -4\\ -36 & 186 & -36 & -45 & -45 & 25 & -45 & -45 & 16 & 25\\ -4 & -36 & 45 & 25 & -36 & -4 & 25 & -36 & 25 & -4\\ -36 & -45 & 25 & 186 & -45 & -36 & -45 & 16 & -45 & 25\\ 25 & -45 & -36 & -45 & 186 & -36 & 16 & -45 & -45 & 25\\ -4 & 25 & -4 & -36 & -36 & 45 & 25 & 25 & -36 & -4\\ -36 & -45 & 25 & -45 & 16 & 25 & 186 & -45 & -45 & -36\\ 25 & -45 & -36 & 16 & -45 & 25 & -45 & 186 & -45 & -36\\ 25 & 16 & 25 & -45 & -45 & -36 & -45 & -45 & 186 & -36\\ -4 & 25 & -4 & 25 & 25 & -4 & -36 & -36 & -36 & 45 \end{array}\right).$$

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