In this answer, @MichaelRozenberg stated the following inequality:
Let $a$, $b$ and $c$ be positive numbers such that $a^3+b^3+c^3=a^2+b^2+c^2.$ Then $$\frac{a}{a^2+2b^2}+\frac{b}{b^2+2c^2}+\frac{c}{c^2+2a^2}\geq 1.$$
Edit: Here is a brute force way:
For $a,b,c>0$ we can prove $$\frac{a}{a^2+2b^2}+\frac{b}{b^2+2c^2}+\frac{c}{c^2+2a^2}\geq\frac{a^2+b^2+c^2}{a^3+b^3+c^3}.$$
Assume WLOG that $a\le b$ and $a\le c$. So $b=a+x$ and $c=a+y$ for some $x,y>0$. Then the above inequality is equivalent to $$42 a^6 \left(x^2-x y+y^2\right)+6 a^5 \left(15 x^3+4 x^2 y-7 x y^2+15 y^3\right)+10 a^4 \left(9 x^4+10 x^3 y-x y^3+9 y^4\right)+a^3 \left(50 x^5+103 x^4 y+34 x^3 y^2+6 x^2 y^3+7 x y^4+50 y^5\right)+a^2 \left(14 x^6+57 x^5 y+27 x^4 y^2+10 x^3 y^3+3 x^2 y^4+9 x y^5+14 y^6\right)+a \left(2 x^7+14 x^6 y+13 x^5 y^2+4 x^4 y^3-4 x^3 y^4+7 x^2 y^5+2 y^7\right)+2 x^3 y \left(x^4+2 x^2 y^2-2 x y^3+y^4\right)\geq0$$ which is true because
- $x^2-xy+y^2=xy+(x-y)^2\geq 0$,
- $15 x^3+4x^2y-7xy^2+15y^3=8(x^3+y^3)+11x^2y+7(x-y)^2(x+y)\geq0$,
- $9 x^4+10 x^3 y - x y^3 +9y^4=8(x^4+y^4) + 11 x^3 y+(x-y)^2(x^2+xy+y^2)$,
- $x^4+2x^2y^2-2xy^3+y^4=x^4+x^2y^2+y^2(x-y)^2\geq0$
I meant the following solution.
By C-S $$\sum_{cyc}\frac{a}{a^2+2b^2}=\sum_{cyc}\frac{a^2(a+2c)^2}{a(a+2c)^2(a^2+2b^2)}\geq\frac{\left(\sum\limits_{cyc}a(a+2c)\right)^2}{\sum\limits_{cyc}a(a+2c)^2(a^2+2b^2)}.$$ Thus, it's enough to prove that $$(a+b+c)^4(a^3+b^3+c^3)\geq(a^2+b^2+c^2)\sum\limits_{cyc}a(a+2c)^2(a^2+2b^2)$$ or $$\sum_{cyc}(4a^6b+3a^5b^2+a^5c^2+a^4b^3-a^4c^3+12a^5bc-8a^4b^2c-4a^4c^2b+8a^3b^3c-16a^3b^2c^2)\geq0$$ or $$\sum_{cyc}(4a^6b+2a^5b^2+2a^4b^3-4a^4b^2c-4a^3b^2c^2)+$$ $$+\sum_{cyc}(a^5b^2+a^5c^2-a^4b^3-a^4c^3+12a^5bc-4a^4b^2c-4a^4c^2b+8a^3b^3c-12a^3b^2c^2)\geq0,$$ which is true by AM-GM and Muirhead.