Htykuut

Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$frac{a^3}{13a^2+5b^2}+frac{b^3}{13b^2+5c^2}+frac{c^3}{13c^2+5a^2}geqfrac{a+b+c}{18}$$

Note that for $(a,b,c)=(0.785, 1.25, 1.861)$, the difference between the LHS and the RHS is around $0.0000158...$.

I tried to use Cauchy-Schwarz, but without success.

Your inequality is equivalent to :
$$sum_{cyc}frac{a}{13}sin(arctan(sqrt{frac{13}{5}}frac{a}{b}))^2geq frac{a+b+c}{18}$$
Each side is divided by $b$

We get :

$$frac{a}{13b} sin(arctan(sqrt{frac{13}{5}}frac{a}{b}))^2+frac{1}{13}sin(arctan(sqrt{frac{13}{5}}frac{b}{c}))^2+frac{c}{13b}sin(arctan(sqrt{frac{13}{5}}frac{c}{a}))^2geq frac{1+frac{a}{b}+frac{c}{b}}{18}$$

Now we put :

$sqrt{frac{13}{5}}frac{a}{b}=x$

$sqrt{frac{13}{5}}frac{b}{c}=y$

$sqrt{frac{13}{5}}frac{c}{a}=z$

Your inequality is equivalent to :

$$sqrt{dfrac{5}{13}}frac{1}{13}dfrac{(x)^3}{(x^2+1)}+dfrac{1}{13}dfrac{(y)^2}{(y^2+1)}+sqrt{dfrac{13}{5}}dfrac{1}{13(y)}dfrac{(z)^2}{(z^2+1)}$$$$geq dfrac{1+sqrt{dfrac{5}{13}}x+sqrt{dfrac{13}{5}}dfrac{1}{y}}{18}$$

With the condition $xyz=(sqrt{frac{13}{5}})^3$

We study the following function :

$$f(x)=sqrt{dfrac{5}{13}}frac{1}{13}dfrac{(x)^3}{(x^2+1)}+dfrac{1}{13}dfrac{(y)^2}{(y^2+1)}+sqrt{dfrac{13}{5}}dfrac{1}{13(y)}dfrac{(z)^2}{(z^2+1)}-dfrac{1+sqrt{dfrac{5}{13}}x+sqrt{dfrac{13}{5}}dfrac{1}{y}}{18}$$

This function is easily differentiable and the minimum is for $x=sqrt{frac{3sqrt{77}}{17}-frac{26}{17}}=alpha$

So with the condition $xyz=(sqrt{frac{13}{5}})^3$ becomes $yz=frac{(sqrt{frac{13}{5}})^3}{alpha}=beta$

So we have this inequality just with $y$ :

$$sqrt{dfrac{5}{13}}frac{1}{13}dfrac{(alpha)^3}{(alpha^2+1)}+dfrac{1}{13}dfrac{(y)^2}{(y^2+1)}+sqrt{dfrac{13}{5}}dfrac{1}{13(y)}dfrac{(frac{beta}{y})^2}{((frac{beta}{y})^2+1)}geqdfrac{1+sqrt{dfrac{5}{13}}alpha+sqrt{dfrac{13}{5}}dfrac{1}{y}}{18}$$

Wich is easily analyzable .

Done !

Second proof :

We have the following theorem :

Let $a,b,c$ be real positive numbers then we have :
$$big(frac{a^3}{13a^2+5b^2}big)^2+big(frac{b^3}{13b^2+5c^2}big)^2+big(frac{c^3}{13c^2+5a^2}big)^2geq frac{a^2+b^2+c^2}{18^2}$$

Proof :

The inequality is equivalent to this :

$$big(frac{1}{13+5x^2}big)^2+x^2big(frac{1}{13+5y^2}big)^2+frac{1}{z^2}big(frac{1}{13+5z^2}big)^2geq frac{1+x^2+frac{1}{z^2}}{18^2}$$

With :

$x=frac{b}{a}$
,
$y=frac{c}{b}$,
$z=frac{a}{c}$

If we put $f(x)=(frac{1}{13+5x})^2$ we have :

$$f(x^2)+x^2f(y^2)+frac{1}{z^2}f(z^2)geq f(1)+f(1)x^2+f(1)frac{1}{z^2}$$

Or :

$$frac{f(x^2)+x^2f(y^2)+frac{1}{z^2}f(z^2)}{1+x^2+frac{1}{z^2}}geq f(1)$$

Now we can apply Jensen's inequality to the function $f(x)$ with the appropriate weight to get :

$$frac{f(x^2)+x^2f(y^2)+frac{1}{z^2}f(z^2)}{1+x^2+frac{1}{z^2}}geq f(frac{x^2+(xy)^2+frac{z^2}{z^2}}{1+x^2+frac{1}{z^2}})=f(1)$$

So it's proved because we have the condition $xyz=1$.

Now it remains to prove the following inequality :

$$sum_{cyc}(frac{a^3}{13a^2+5b^2})(frac{b^3}{13b^2+5c^2})geq frac{ab+bc+ca}{18^2}$$

I put it in a question because I can't solve it now .