Prove that $\sum_{\mathrm{cyc}} \frac{214x^4}{133x^3 + 81y^3} \ge x + y + z$ for $x, y, z > 0$

Question

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.

Answer 1

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.

Answer 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.

Answer 3

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

Answer 4

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 $y\ge0$