# 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). – Adam P. Goucher Jan 12 '15 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. – John Bentin Jan 12 '15 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). – Todd Trimble Jan 12 '15 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..! – Malik Younsi Jan 12 '15 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. – Yaakov Baruch Jan 16 '15 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. – math110 Jan 13 '15 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... – Joonas Ilmavirta Jan 13 '15 at 17:41

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 $$S_3S_1 \ge S_2^2$$.