An inequality with a prime number

Question

I have this to purpose :

Let $a,b,c$ be real positive numbers such that $abc=1$ prove that : $$\frac{a^2}{(a^{11}+1)^2}+\frac{b^2}{(b^{11}+1)^2}+\frac{c^2}{(c^{11}+1)^2}\leq \frac{10^7}{9193531}$$

Where $9193531$ is a prime number wich makes the problem harder .

Really I have no ideas to prove this and all my classical methods fails automatically when I want solve this problem .

Furthermore I do not know which class of functions belongs $f(x)$ where :

$$f(x)=\frac{x^2}{(x^{11}+1)^2}$$

Maybe we can find a way with the concept of Quasiconvex function

If you have any hints it would be nice.

Thanks

Answer 1

Let $$f(x)=\frac{x^2}{(x^{11}+1)^2}.$$ By a compactness argument, there exists a triple $(a,b,c)$ of positive real numbers that maximizes the sum $f(a)+f(b)+f(c)$ subject to the constraint $abc=1$. We may assume that $0<a\leq b\leq c$. Then $a\leq1$ and $c\geq1$. If $a<10^{-1/11}$ then increasing $a$ and decreasing $c$ contradicts the maximality of $(a,b,c)$. Thus, $a\geq10^{-1/11}$. If $b>1$ then $$f(a)+f(b)+f(c)\leq f(10^{-1/11})+f(1)+f(1)\leq1.044$$ which contradicts the maximality of $(a,b,c)$ (there are better solutions than $1.044$). Thus, $b\leq1$. In summary, $$10^{-1/11}\leq a\leq b\leq1\leq c.$$ By Lagrange multipliers, $$\left\langle f^\prime(a),f^\prime(b),f^\prime(c)\right\rangle=\lambda\left\langle bc,ac,ab\right\rangle=\lambda\left\langle\frac{1}{a},\frac{1}{b},\frac{1}{c}\right\rangle.$$ If we set $p(x)=xf^\prime(x)$ then $p(a)=p(b)=p(c)$. Since $a\leq b\leq1$, looking at the graph of $p(x)$ shows that $a=b$. Then we wish to maximize the one-variable function $$g(x)=2f(x)+f\left(\frac{1}{x^2}\right)=2\left(\frac{x}{x^{11}+1}\right)^2+\left(\frac{x^{20}}{x^{22}+1}\right)^2.$$ The maximum occurs slightly above $x=10^{-1/11}$. We can find the maximum by setting $g^\prime(x)=0$. We can compute $$g^\prime(x)=\frac{4x(1-10x^{11})}{(x^{11}+1)^3}+\frac{4x^{39}(10-x^{22})}{(x^{22}+1)^3}.$$ Setting $g^\prime(x)=0$ gives $$\frac{1-10x^{11}}{(x^{11}+1)^3}+\frac{x^{38}(10-x^{22})}{(x^{22}+1)^3}=0.$$ Clearing denominators gives $$(1-10x^{11})(x^{22}+1)^3+x^{38}(10-x^{22})(x^{11}+1)^3=0.$$ The root of this polynomial can be found to arbitrary precision using Newton's method: $$x\approx0.8114699441003698687884808010408458056245.$$ Then $$g(x)\approx1.0877212680663091337501932514101994261908490$$ which is about $0.0000000873$ below $10^7/9193531$.

To make this 100% rigorous, you need to use bounds on $g^\prime(x)$ (this is rather tedious and was done in an earlier version of this answer).

Answer 2

Geometric Explanation "min/max at x=y"

(Not an Answer)

Using $c=1/(ab)$, Let $a=x$, $b=y$ and define: $$ \begin{align} &f(x,y)=\frac{10^7}{9193531}-\frac{x^{2}}{\left(x^{11}+1\right)^2}-\frac{y^{2}}{\left(y^{11}+1\right)^2}-\frac{(xy)^{20}}{\left((xy)^{11}+1\right)^2}=f(y,x) \\ &f({\small-\frac{8}{10}},{\small-\frac{9}{10}})\approx-1.4\lt0, \quad f({\small2},{\small2})\approx+1.0\gt0 \end{align} $$

Benefiting from the function symmetric behavior $f(x,y)=f(y,x)$, the inequality holds for the special case min/max accord at $x=y$. Thus, re-define: $$ \begin{align} &f(x)=\frac{10^7}{9193531}-\frac{2x^{2}}{\left(x^{11}+1\right)^2}-\frac{x^{40}}{\left(x^{22}+1\right)^2} \\ &\qquad=\color{blue}{\frac{10^7}{9193531}-\frac{2x^{2}+4x^{24}+x^{40}+2x^{46}+2x^{51}+x^{62}}{1+2x^{11}+3x^{22}+4x^{33}+3x^{44}+2x^{55}+x^{66}}} \end{align} $$ And the question now turns to be about $\,{\small f(x)}\,$ roots for $\,{\small x\gt0}\,$.



Putting $\,{\small f(x)=0}\,$ and simplifying, gives the equivalent polynomial: $$ \begin{align} p(x)=\,&\,\color{red}{{10^7}\left({1+2x^{11}+3x^{22}+4x^{33}+3x^{44}+2x^{55}+x^{66}}\right)} \\ &\,\color{red}{-{9193531}\left(2x^{2}+4x^{24}+x^{40}+2x^{46}+2x^{51}+x^{62}\right)} \end{align} $$ Because the polynomial degree is high, and to determine if $\,{\small p(x)}\,$ has any $\,{\color{red}{\small\text{real root}}}\,$ for $\,{\small x\gt0}\,$, it is enough to use Sturm theorem to compare number of roots inside the two regions $\,{\small [-1,0]}\,$ and $\,{\small [-1,\infty)}\,$, which should result in only $\,{\color{red}{\small\text{ONE}}}\,$ root in both, hence $\,{\small p(x)\ne0\,\,\colon\,x\in(0,\infty)}\,$.