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Problem. Let $x, y, z > 0$. Prove that $$\frac{214x^4}{133x^3 + 81y^3} + \frac{214y^4}{133y^3 + 81z^3} + \frac{214z^4}{133z^3 + 81x^3} \ge x+y+z.$$

It is verified by Mathematica. The inequality holds with equality if $x = y = z$. When $x = \frac{121}{84}, y = \frac{43}{66}$ and $z = 1$, $\mathrm{LHS} - \mathrm{RHS} \approx 0.000005327884220$.

It is a stronger version of the inequality in this link: Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$

They can be written as (for $k = \frac{8}{5}$ and $k = \frac{133}{81} \approx 1.641975$, respectively) $$\frac{x^4}{kx^3+y^3} + \frac{y^4}{ky^3 + z^3} + \frac{z^4}{kz^3 + x^3} \ge \frac{x+y+z}{k+1}.$$

The best constant $k$ is approximately $1.64199$ (see the comment by @Colescu in the link above).

I can prove the inequality of $k = \frac{8}{5}$ by the Buffalo Way. Several months ago, I tried to prove the inequality of $k = \frac{133}{81}$ by the Buffalo Way without success. However, I think that the Buffalo Way may work but just I have not found the way.

Any comments and solutions are welcome.

Michael Rozenberg
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River Li
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4 Answers4

4

We can reduce a degree of this inequality.

Indeed, by C-S we obtain: $$\sum_{cyc}\frac{x^4}{133x^3+81y^3}=\sum_{cyc}\frac{x^4(200x-57y+154z)^2}{(133x^3+81y^3)(200x-57y+154z)^2}\geq$$ $$\geq\frac{\left(\sum\limits_{cyc}(200x^3-57x^2y+154x^2z)\right)^2}{\sum\limits_{cyc}(133x^3+81y^3)(200x-57y+154z)^2}$$ and it's enough to prove that: $$214\left(\sum\limits_{cyc}(200x^3-57x^2y+154x^2z)\right)^2\geq(x+y+z)\sum\limits_{cyc}(133x^3+81y^3)(200x-57y+154z)^2,$$ which is true, but BW does not help here and I have no a proof by hand.

Michael Rozenberg
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    (+1) You found the key. Let $z = \min(x,y,z)$. If $x \ge y \ge z$, BW works. – River Li Jun 07 '20 at 16:12
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    @River Li For $x\geq y\geq z$ it's obvious. The hardest case it's $x\leq y\leq z$. – Michael Rozenberg Jun 07 '20 at 16:28
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    Your approach works for larger $k$ by designing appropriate coefficients of $ax+by+cz$, right? – River Li Jun 08 '20 at 00:56
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    @River Li Yes. It's interesting that we can get a best estimation by using C-S with a linear bacteria $x+py+qz$. It's not always possible! – Michael Rozenberg Jun 08 '20 at 01:02
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    Yes, it's amazing. For some inequality problems, we need to find quadratic or higher polynomials which is difficult to determined, e.g., https://math.stackexchange.com/questions/1793680/prove-sum-limits-cyc-left-fraca4a3b3-right-frac34-geqslant-fr/1807882#comment7618834_1807882 – River Li Jun 08 '20 at 01:09
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    I think the last one is hard to solve by SOS, we get$:$ $\text{LHS}-\text{RHS}=\sum\limits_{cyc} \left( {x}^{4}m_{{1}}+{x}^{3}ym_{{8}}+{x}^{3}zm_{{9}}+{x}^{2}{y}^{2}m {{6}}+{x}^{2}yzm{{15}}+{x}^{2}{z}^{2}m_{{7}}+x{y}^{3}m_{{4}}+x{y}^{2 }zm_{{14}}+xy{z}^{2}m_{{13}}+x{z}^{3}m_{{5}}+{y}^{4}m_{{2}}+{y}^{3}zm_ {{12}}+{y}^{2}{z}^{2}m_{{11}}+y{z}^{3}m_{{10}}+{z}^{4}m_{{3}} \right) \left( x-y \right) ^{2}.$ Get an equation give https://imgur.com/gz0UvDe.

    So I wish to know your proof.

     – NKellira Jun 08 '20 at 11:29
  • @tthnew Thanks. I have not yet get a proof. – River Li Aug 15 '20 at 03:44
  • @RiverLi I also no. – NKellira Aug 17 '20 at 09:30
  • @MichaelRozenberg how you you choose $p, q$ to get the true inequality$?$ It's very good. – NKellira Aug 17 '20 at 09:31
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    @tthnew I applied the point $(x,y,z)=\left(\frac{121}{84},\frac{43}{66},1\right)$ and solved a system. I used that in C-S the equality occurs, when vectors are parallel. – Michael Rozenberg Aug 17 '20 at 09:36
  • The point you choose is nice, and you solution also. Still now I can not prove your last inequality. – NKellira Aug 17 '20 at 09:56
  • Can we take the multiplying factor to be $52x-23y+29z$ instead of $200x-57y+154z$? If we take $x=\min{x,y,z}$, the $x\le z\le y$ case is obvious and for the $x\le y \le z$ case, we get $574144(u^2+uv+v^2)x^4 + (2206885u^3-1321290u^2v-3348793uv^2+ 89691v^3)x^3 + (3597995u^4-2067245u^3v-9317913u^2v^2-3652673uv^3+422204v^4)x^2 + (2267473u^5+429921u^4v-4493731u^3v^2-2538900u^2v^3+1326155uv^4+604438v^5)x + 478394u^6+566165u^5v+31078u^4v^2+800983u^3v^3+1950293u^2v^4+1092617uv^5+176175v^6$ – maiar Jun 01 '22 at 09:31
  • @MichaelRozenberg Have you prove the last inequality of your answer? – TATA box Dec 06 '23 at 11:24
2

Here I present the computer-generated Buffalo Way solution. I didn't think the solution to be this ugly when I coded this, but it happens.

We want to prove $$872613 x^{7}y^{3} + 531441 x^{7}z^{3} + 872613 x^{6}y^{4} - 1432809 x^{6}y^{3}z - 872613 x^{6}yz^{3} - 872613 x^{6}z^{4} - 872613 x^{4}y^{6} + 901368 x^{4}y^{3}z^{3} + 872613 x^{4}z^{6} + 531441 x^{3}y^{7} - 872613 x^{3}y^{6}z + 901368 x^{3}y^{4}z^{3} + 901368 x^{3}y^{3}z^{4} - 1432809 x^{3}yz^{6} + 872613 x^{3}z^{7} - 1432809 xy^{6}z^{3} - 872613 xy^{3}z^{6} + 872613 y^{7}z^{3} + 872613 y^{6}z^{4} - 872613 y^{4}z^{6} + 531441 y^{3}z^{7}$$ is positive.

The Buffalo Way method helps.

This polynomial is symmetric.

Let $x = \min\{x, y, z\}$, $y = x + a$, $z = x + b$.

Substitution gives $$\left(3016116 a^{2} - 3016116 a b + 3016116 b^{2}\right) x^{8} + \left(7540290 a^{3} + 24659883 a^{2} b - 23151825 a b^{2} + 7540290 b^{3}\right) x^{7} + \left(15461928 a^{4} + 51247485 a^{3} b + 25272972 a^{2} b^{2} - 60313167 a b^{3} + 15461928 b^{4}\right) x^{6} + \left(15652602 a^{5} + 63883971 a^{4} b + 75012399 a^{3} b^{2} - 33150789 a^{2} b^{3} - 49375413 a b^{4} + 15652602 b^{5}\right) x^{5} + \left(7522956 a^{6} + 44891091 a^{5} b + 95298039 a^{4} b^{2} + 11145762 a^{3} b^{3} - 46208151 a^{2} b^{4} - 11765817 a b^{5} + 7522956 b^{6}\right) x^{4} + \left(1404054 a^{7} + 16644285 a^{6} b + 60598125 a^{5} b^{2} + 44855208 a^{4} b^{3} - 28491912 a^{3} b^{4} - 13640319 a^{2} b^{5} + 3619161 a b^{6} + 1404054 b^{7}\right) x^{3} + \left(2617839 a^{7} b + 19262124 a^{6} b^{2} + 30670731 a^{5} b^{3} - 15018129 a^{3} b^{5} + 3306744 a^{2} b^{6} + 1594323 a b^{7}\right) x^{2} + \left(2617839 a^{7} b^{2} + 8165934 a^{6} b^{3} + 5235678 a^{5} b^{4} - 5235678 a^{4} b^{5} - 642978 a^{3} b^{6} + 1594323 a^{2} b^{7}\right) x + 872613 a^{7} b^{3} + 872613 a^{6} b^{4} - 872613 a^{4} b^{6} + 531441 a^{3} b^{7}$$

If we let $X = \frac{x}{\sqrt{a}\sqrt{b}}$,

We split into $58$ cases.

(1) For ${a}/{b}$ in range $\left[0.0, 0.4304\right]$, the polynomial is larger than $$\left(5289724 X^{8} + 9721601 X^{7} - 6472990 X^{6} - 27945512 X^{5} - 15542184 X^{4} + 13520031 X^{3} + 19931661 X^{2} + 8919859 X + 1378640\right)\left(\sqrt{a}\sqrt{b}\right)^{10}$$, which is positive.

(2) For ${a}/{b}$ in range $\left[0.4304, 0.4338\right]$, the polynomial is larger than $$\left(5245055 X^{8} + 9635887 X^{7} - 6470872 X^{6} - 27446186 X^{5} - 15442001 X^{4} + 13309502 X^{3} + 19625777 X^{2} + 8769605 X + 1355267\right)\left(\sqrt{a}\sqrt{b}\right)^{10}$$, which is positive.

(3) For ${a}/{b}$ in range $\left[0.4338, 0.4356\right]$, the polynomial is larger than $$\left(5221753 X^{8} + 9592263 X^{7} - 6456145 X^{6} - 27258913 X^{5} - 15339842 X^{4} + 13209802 X^{3} + 19470823 X^{2} + 8692567 X + 1343224\right)\left(\sqrt{a}\sqrt{b}\right)^{10}$$, which is positive.

(4) For ${a}/{b}$ in range $\left[0.4356, 0.4367\right]$, the polynomial is larger than $$\left(5207630 X^{8} + 9566190 X^{7} - 6442823 X^{6} - 27149048 X^{5} - 15272265 X^{4} + 13152771 X^{3} + 19378452 X^{2} + 8646323 X + 1335975\right)\left(\sqrt{a}\sqrt{b}\right)^{10}$$, which is positive.

The 5-58 cases are put in the link below. https://sagecell.sagemath.org/?q=vckhdl

Therefore we conclude that the polynomial is positive.

River Li
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didgogns
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0

Hard case $x\leq y\leq z$ and $\frac{y}{x},\frac{z}{y}\in[\frac{121}{84},\infty)$:

We have the function :

$$h\left(x\right)=\frac{13}{8+5x^{3}}+\frac{83}{1000}\left(1-\frac{2x}{2x+1}\right)+\frac{0.2375x}{x+1}$$

And :

$$g\left(x\right)=\frac{214}{133+81x^{3}}$$

Lemma :

And for $x\in[\frac{121}{84},\infty)$ a real number :

$$f''(x)=g''(x)-h''(x)>0$$

The problem :

The inequality is equivalent to :

$$\frac{x}{x+y}g\left(\frac{y}{x}\right)+\frac{y}{x+y}g\left(\frac{z}{y}\right)+\frac{1}{x+y}\frac{214z^{4}}{133z^{3}+81x^{3}}-\frac{\left(x+y+z\right)}{x+y}\geq 0$$

Now we use the lemma and Jensen's inequality we got ($f(x)=g(x)-h(x)$):

$$\frac{x}{x+y}f\left(\frac{y}{x}\right)+\frac{y}{x+y}f\left(\frac{z}{y}\right)+\frac{1}{x+y}\frac{214z^{4}}{133z^{3}+81x^{3}}-\frac{\left(x+y+z\right)}{x+y}\geq f\left(\frac{\left(z+y\right)}{x+y}\right)+\frac{1}{x+y}\frac{214z^{4}}{133z^{3}+81x^{3}}-\frac{\left(x+y+z\right)}{x+y}$$

Or after summation:

$$a(x,y,z)=\frac{x}{x+y}h\left(\frac{y}{x}\right)+\frac{y}{x+y}h\left(\frac{z}{y}\right)+\frac{1}{x+y}\frac{214z^{4}}{133z^{3}+81x^{3}}+g\left(\frac{\left(z+y\right)}{x+y}\right)-h\left(\frac{\left(z+y\right)}{x+y}\right)-\frac{\left(x+y+z\right)}{x+y}\geq 0$$

Remains to show that it is positive and BW helps but it's lenghty and I cannot check all the coefficients .

We have as expression expanded $y\geq 1$ a real number:

$$a(1,1+x,1+xy)= \mbox{the very long expression is given in the link blow}$$https://sagecell.sagemath.org/?q=ssnesu

Edit 18/04/2022:

Let $x\ge1 $ then it seems we have :

$$g''\left(x\right)-h''\left(x\right)-\frac{17}{100}f''\left(x\right)>0$$

Where :

$$f\left(x\right)=\frac{214x^{-1}}{133+81x^{3}}+\frac{x^{2}}{x}\frac{214}{133+81x^{-3}}-\frac{13x^{-1}}{8+5x^{3}}-\frac{x^{2}}{x}\frac{13}{8+5x^{-3}}$$

And :

$$h\left(x\right)=\frac{13}{8+5x^{3}}+\frac{0.13x}{x+1}+\frac{17}{100}f\left(x\right)$$

And :

$$g\left(x\right)=\frac{214}{133+81x^{3}}$$

So starting from :

$$b(x,y,z)=\frac{x}{x+y}K\left(\frac{y}{x}\right)+\frac{y}{x+y}K\left(\frac{z}{y}\right)+\frac{1}{x+y}\frac{214z^{4}}{133z^{3}+81x^{3}}+g\left(\frac{\left(z+y\right)}{x+y}\right)-K\left(\frac{\left(z+y\right)}{x+y}\right)-\frac{\left(x+y+z\right)}{x+y}\geq 0$$

Where :

$$K(x)=h(x)+0.17f(x)$$

We can use Buffalo's way but it's not for the hand .

Edit :

Using tangent line method on $r(\frac{\left(z+y\right)}{x+y})=g\left(\frac{\left(z+y\right)}{x+y}\right)-h\left(\frac{\left(z+y\right)}{x+y}\right)$,$q(x)=-0.027126(x-((((121)/(84))+1)/(((43)/(66))+1)))-((15865219)/(100000000))<r\left(x\right)$ , $a=(121/84+1)/(43/66+1)$ we have for $\frac{\left(z+y\right)}{x+y}\geq a$ :

$$a(x,y,z)=\frac{x}{x+y}h\left(\frac{y}{x}\right)+\frac{y}{x+y}h\left(\frac{z}{y}\right)+\frac{1}{x+y}\frac{214z^{4}}{133z^{3}+81x^{3}}+q\left(\frac{\left(z+y\right)}{x+y}\right)-\frac{\left(x+y+z\right)}{x+y}$$

$$a(1,1+x/2,1+((x y)/(y+1)))=\mbox{the very long expression is given in the link below}$$https://sagecell.sagemath.org/?q=gprkws

River Li
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Miss and Mister cassoulet char
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  • For some details on the second derivative see WA . – Miss and Mister cassoulet char Apr 15 '22 at 09:24
  • @RiverLi What do think about ? – Miss and Mister cassoulet char Apr 15 '22 at 09:25
  • Do you have a full proof? If so, please give all details (step-by-step). If you just give partial result, I will wait for a full proof (perhaps in the future, you can do it). – River Li Apr 15 '22 at 23:17
  • Sorry, It is difficult for me to follow your current writing. Remark: For a full step-by-step proof, there are no sentences "Now it's really easy ." "Can you end it now?" "The rest is smooth." – River Li Apr 15 '22 at 23:19
  • @RiverLi Sorry I cannot do more but some explanation : You have already an half proof with the Rozenberg's answer . On the other hand $a(x,y,z)$ is the consequence of using Jensens' inequality (divide by $(x+y)$ , and use $ \frac{1}{x+y}\frac{214x^{4}}{133x^{3}+81y^{3}}=\frac{x}{x+y}\frac{214}{133+81\left(\frac{y}{x}\right)^{3}}$ in your inequality . – Miss and Mister cassoulet char Apr 16 '22 at 14:33
  • @RiverLi Again I haven't the time but I think you can end up by yourself :-). – Miss and Mister cassoulet char Apr 16 '22 at 14:35
  • Your writing is hard to read. Sorry. I don't know how $a(x, y, z)$ is related to OP's inequality. You should write details. If you don't have time, in the future (perhaps 1 or 2 years), you can complete it. – River Li Apr 16 '22 at 15:42
  • @RiverLi Hum well ok but it's the last time . – Miss and Mister cassoulet char Apr 16 '22 at 15:56
  • OK. See you 1-2 years for the full proof of OP's inequality. – River Li Apr 16 '22 at 16:03
  • @RiverLi Now we have the hard part . – Miss and Mister cassoulet char Apr 16 '22 at 17:15
  • You don't prove this case even, right? You said "Remains to show that it is positive and BW helps but it's lenghty and I cannot check all the coefficients ." – River Li Apr 16 '22 at 22:36
  • In the future, when you have a full proof for $x\leq y\leq z$ (that is, NO $\frac{y}{x},\frac{z}{y}\in[\frac{121}{84},\infty)$), I will check your proof, OK? – River Li Apr 16 '22 at 22:41
  • By the way, I did not downvote you. – River Li Apr 19 '22 at 00:43
  • @RiverLi I found a proof but it's really for a computer . Try and keep me informed that you think about . – Miss and Mister cassoulet char Apr 20 '22 at 13:32
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    Dear Erik Satie, don't give a hint. You should give step-by-step proof. Perhaps you stop for proving this difficult inequality. There are many inequalities waiting for you. For example, please add details (general $n$) for this answer https://math.stackexchange.com/questions/4351612/prove-that-displaystyle-prod-k-1n-frac1x-kx-k-geq-prod-k-1n-fr/4399286#4399286. It is difficult to prove general $n$. If you can't prove the general $n$, it is nothing. (If you indeed don't have a proof for general $n$, please add information "I don't have a proof for general $n$. I only prove $n=3$.) – River Li Apr 20 '22 at 13:38
0

My first partial answer :

We introduce the function :

$$f\left(x\right)=\frac{1}{133+81x^{6}}$$

The function is convex for $x\in[1.03,\infty)$ so we have using Jensen's inequality :

$$g\left(x,y,z\right)=\left(x^{2}+0.001\right)f\left(\frac{\left(0.001\frac{z}{y}+xy\right)}{x^{2}+0.001}\right)+0.999f\left(\frac{z}{y}\right)+\frac{z^{8}}{133z^{6}+81x^{6}}-\frac{\left(x^{2}+y^{2}+z^{2}\right)}{214}\geq^?0$$

Using Buffalo's way we have :

$$g(x,1,(1+x^2)(1+y))= \mbox{the very very long expression is given in the link below} \ge 0$$https://sagecell.sagemath.org/?q=gmrshy

Where I only set the numerator For $x\in[0,1]$ and $y\ge0$

River Li
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Miss and Mister cassoulet char
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