This inequality why can't solve it by now (Only four variables inequality)?

Question

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 $$\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{d}+\dfrac{d^2}{a}\right)^4-64(a^2+b^2+c^2+d^2)=\dfrac{a^{12}c^4d^4+4a^{10}b^3c^3d^4+4a^{10}bc^6d^3+4a^9bc^4d^6 +6a^8b^6c^2d^4+12a^8b^4c^5d^3+64a^8b^4c^4d^4+6a^8b^2c^8d^2+12a^7b^4c^3d^6+12a^7b^2c^6d^5+4a^6b^9cd^4+\cdots+4ab^4c^6d^9+b^4c^4d^{12}}{a^4b^4c^4d^4}$$

Answer 1

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 $(a,b,c,d)\in S$.

Answer 2

$\newcommand{\res}{\operatorname{res}}$ 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). $\Box$

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

Answer 3

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.

Answer 4

This is not an answer. I want to point out two possible directions. Refer to this link https://math.stackexchange.com/questions/1829520/prove-this-by-inequality-with-four-variables-inequality

1) We may consider the following problem: Prove (or disprove) that \begin{align} \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]. \end{align} This is based on the following result: \begin{align} &\Big(4(a^2+b^2+c^2+d^2)+\frac{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 ($Q$ is positive semidefinite) $$z = \left(\begin{array}{c} { {a}}^2\\ {a} {b}\\ { {b}}^2\\ {a} {c}\\ {b} {c}\\ { {c}}^2\\ {a} {d}\\ {b} {d}\\ {c} {d}\\ { {d}}^2 \end{array}\right), \quad Q = \left(\begin{array}{cccccccccc} 45 & -36 & -4 & -36 & 25 & -4 & -36 & 25 & 25 & -4\\ -36 & 186 & -36 & -45 & -45 & 25 & -45 & -45 & 16 & 25\\ -4 & -36 & 45 & 25 & -36 & -4 & 25 & -36 & 25 & -4\\ -36 & -45 & 25 & 186 & -45 & -36 & -45 & 16 & -45 & 25\\ 25 & -45 & -36 & -45 & 186 & -36 & 16 & -45 & -45 & 25\\ -4 & 25 & -4 & -36 & -36 & 45 & 25 & 25 & -36 & -4\\ -36 & -45 & 25 & -45 & 16 & 25 & 186 & -45 & -45 & -36\\ 25 & -45 & -36 & 16 & -45 & 25 & -45 & 186 & -45 & -36\\ 25 & 16 & 25 & -45 & -45 & -36 & -45 & -45 & 186 & -36\\ -4 & 25 & -4 & 25 & 25 & -4 & -36 & -36 & -36 & 45 \end{array}\right).$$

2) We may consider the following problem: Prove (or disprove) that $$\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a} \ge 4 S_1 + \frac{4(S_4 - S_1^4)}{S_3 + 2S_2S_1 + S_1^3}$$ where $$S_1 = \frac{a+b+c+d}{4}, \ S_2 = \frac{a^2+b^2+c^2+d^2}{4}, \ S_3 = \frac{a^3+b^3+c^3+d^3}{4}, \ S_4 = \frac{a^4+b^4+c^4+d^4}{4}.$$ This is based on the following result: $$4 S_1 + \frac{4(S_4 - S_1^4)}{S_3 + 2S_2S_1 + S_1^3} \ge 4\sqrt[4]{\frac{a^4+b^4+c^4+d^4}{4}}$$ which follows from the facts: $$\sqrt[4]{S_4} = S_1 + \frac{\sqrt{S_4} - S_1^2}{\sqrt[4]{S_4} + S_1} = S_1 + \frac{S_4 - S_1^4}{(\sqrt[4]{S_4} + S_1)(\sqrt{S_4} + S_1^2)}$$ and \begin{align} (\sqrt[4]{S_4} + S_1)(\sqrt{S_4} + S_1^2) &\ge (\sqrt[3]{S_3} + S_1) (\sqrt[3]{S_3^2} + S_1^2)\nonumber\\ &= S_3 + S_1\sqrt[3]{S_3^2} + S_1^2\sqrt[3]{S_3} + S_1^3 \nonumber\\ &\ge S_3 + 2S_1 \sqrt{S_3S_1} + S_1^3 \nonumber\\ &\ge S_3 + 2S_1S_2 + S_1^3 \end{align} where we have used the power mean inequality to obtain $\sqrt[4]{S_4} \ge \sqrt[3]{S_3} \ge \sqrt{S_2} \ge S_1$, and the Cauchy-Bunyakovsky-Schwarz inequality to obtain $S_3S_1 \ge S_2^2$.