Inequality $\frac{a}{a^2+b^2}+\frac{b}{b^2+c^2}+\frac{c}{c^2+a^2}\geq 1.5$ with a condition

Question

It's a problem create by my own

Let $a,b,c>0$ such that $a^3+b^3+c^3=a^2+b^2+c^2$ then we have :
$$\frac{a}{a^2+b^2}+\frac{b}{b^2+c^2}+\frac{c}{c^2+a^2}\geq 1.5$$

I try to get an homogeneous inequality because we have :

$$\frac{a}{a^2+b^2}+\frac{b}{b^2+c^2}+\frac{c}{c^2+a^2}\geq 1.5\frac{a^2+b^2+c^2}{a^3+b^3+c^3}$$

But the condition stop me (it's not homogeneous )

I try to use the Gauss's identity :

$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$

But it reveals nothing good .

If we multiply by $(a^2+b^2)(b^2+c^2)(c^2+a^2)$ we get :

$$(a^2 + a b + a c + b^2 + b c + c^2) (a^2 b – a b c + a c^2 + b^2 c)\geq 1.5(a^2+b^2)(b^2+c^2)(c^2+a^2)$$

But I'm stuck now…

If you have nice ideas it would be nice

Thanks a lot for sharing your time and knowledge .

Answer 1

It's wrong.

After homogenization try $b=1$ and $c\rightarrow0^+$.

We obtain: $$2a^5-a^4+2a^3-4a^2+2a-1\geq0,$$ which is not so true.

The following inequality is true already.

Let $a$, $b$ and $c$ be positive numbers such that $a^3+b^3+c^3=a^2+b^2+c^2.$ Prove that: $$\frac{a}{a^2+2b^2}+\frac{b}{b^2+2c^2}+\frac{c}{c^2+2a^2}\geq 1.$$