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I would like to solve this

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

We get :

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

Now we put :

$\sqrt{\frac{13}{5}}\frac{a}{b}=\frac{A}{B}$

$\sqrt{\frac{13}{5}}\frac{b}{c}=\frac{B}{C}$

$\sqrt{\frac{13}{5}}\frac{c}{a}=\frac{C}{A}$

We find :

$$\sqrt{\frac{5}{13}}\frac{A}{13B}\sin(\arctan(\frac{A}{B}))^2+\frac{1}{13}\sin(\arctan(\frac{B}{C}))^2+\sqrt{\frac{13}{5}}\frac{C}{13B}\sin(\arctan(\frac{C}{A}))^2\geq \frac{1+\sqrt{\frac{5}{13}}\frac{A}{B}+\sqrt{\frac{13}{5}}\frac{C}{B}}{18}$$

After that we put :

$\frac{A}{B}=\frac{x+y}{1-xy}$

$\frac{B}{C}=\frac{z+y}{1-zy}$

$\frac{C}{A}=\frac{x+z}{1-xz}$

With $xy<1$ , $yz<1$ et $zx<1$

Now we use the following identity :

$\arctan(x)+\arctan(y)=\arctan(\frac{x+y}{1-xy})$

We find :

$\sqrt{\frac{5}{13}}\frac{x+y}{13(1-xy)}\sin(\arctan(x)+\arctan(y))^2+\frac{1}{13}\sin(\arctan(z)+\arctan(y))^2+\sqrt{\frac{13}{5}}\frac{1-zy}{13(z+y)}\sin(\arctan(x)+\arctan(z))^2$

$\geq \frac{1+\sqrt{\frac{5}{13}}\frac{x+y}{1-xy}+\sqrt{\frac{13}{5}}\frac{1-zy}{y+z}}{18}$

Moreover $\sin(\arctan(u)+\arctan(v))^2=\frac{(u+v)^2}{(u^2+1)(v^2+1)}$

So we get the following inequality :

$\sqrt{\frac{5}{13}}\frac{x+y}{13(1-xy)}\frac{(x+y)^2}{(x^2+1)(y^2+1)}+\frac{1}{13}\frac{(z+y)^2}{(z^2+1)(y^2+1)}+\sqrt{\frac{13}{5}}\frac{1-zy}{13(z+y)}\frac{(x+z)^2}{(x^2+1)(z^2+1)}\geq \frac{1+\sqrt{\frac{5}{13}}\frac{x+y}{1-xy}+\sqrt{\frac{13}{5}}\frac{1-zy}{y+z}}{18}$

After that I think we can use trigonometric or hyperbolic properties but I don't know how .

1 Answers1

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So, my attempt is based on putting: $\tfrac{b}{a}=x$ and $\tfrac{c}{b}=y$, so $x,y>0$. Then our OP inequality: $$\frac{a^3}{13a^2+5b^2}+\frac{b^3}{13b^2+5c^2}+\frac{c^3}{13c^2+5a^2}\geq\frac{a+b+c}{18}$$ $$\frac{a}{13+5x^2}+\frac{b}{13+5y^2}+\frac{c}{13+5(\tfrac{1}{xy})^2}\geq\frac{a+b+c}{18}$$ $$\frac{\tfrac{b}{x}}{13+5x^2}+\frac{b}{13+5y^2}+\frac{yb}{13+5(\tfrac{1}{xy})^2}\geq\frac{\tfrac{b}{x}+b+yb}{18}$$ $$\frac{\tfrac{1}{x}}{13+5x^2}+\frac{1}{13+5y^2}+\frac{y}{13+5(\tfrac{1}{xy})^2}\geq\frac{\tfrac{1}{x}+1+y}{18}$$ $$\frac{1}{13+5x^2}+\frac{x}{13+5y^2}+\frac{(xy)^3}{13(xy)^2+5}\geq\frac{1+x+xy}{18}$$ $$\frac{1}{13+5x^2}+\frac{x}{13+5y^2}+\frac{(xy)^3}{13(xy)^2+5}-\frac{1+x+xy}{18}\geq0$$ So, it's enough to study 2-variables inequality, but I can't continue it in terms of elementary inequalities, by the way wolf says

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