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

• I don't think this is research-level, since it can (if true) be proved by converting to an equivalent polynomial inequality and appealing to Charles Delzell's explicit algorithm for expressing any positive-semidefinite polynomial as a sum of squares of rational functions (c.f. Hilbert's seventeenth problem). Jan 12, 2015 at 17:17
• @AdamP.Goucher: The polynomial you refer to is of the twentieth degree, and some of its integer coefficients are in the hundreds. Delzell's algorithm might be rather laborious in this case, unless we entrust the job to a computer. Jan 12, 2015 at 17:45
• Just to clarify: would you accept documentation of a computer certification as an answer, or do you want only a "human" proof? To me it looks like the question could stand (as an MO question). Jan 12, 2015 at 17:57
• @AdamP.Goucher : But in your book "Mathematical Olympiad Dark Arts", you say "Hence [by Delzell's algorithm] it is theoretically possible to prove any inequality involving rational functions simply by reducing it to the sum of squares inequality. However, this approach is similar in its impracticality to building an automobile using Stone Age tools." As for the present problem, I would be quite interested to see the automobile built more efficiently..! Jan 12, 2015 at 18:35
• I checked empirically that the exponents 4 on the RHS of the inequality can be increased to about 4.8. The quadratic form corresponding to 2nd derivatives around (1, 1, 1, 1) is positive definite for exponents $\le$ 5. Jan 16, 2015 at 14:37

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

• Hello,your idea is nice,But I think following works is key. Jan 13, 2015 at 16:14
• @math110, if this line of attack ever leads to a full proof, the calculations I have done are clearly only one of at least two keys. Using Lagrange's theorem immediately removes all roots from the problem, and the observation $f=4\lambda$ simplifies the problem further, but one still has to show that $a=b=c=d=1$ is the only solution to the resulting equation. At least it's in principle possible to work with equations instead of inequalities after applying Lagrange's theorem... Jan 13, 2015 at 17:41

$$\newcommand{\res}{\operatorname{res}}$$ 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 $$b and $$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):

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.

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 $$S_3S_1 \ge S_2^2$$.