Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=ab+ac+bc$. Prove that: $$\frac{a}{a^2+2b}+\frac{b}{b^2+2c}+\frac{c}{c^2+2a}\leq1$$
We can try the following.
Replace $a\rightarrow\frac{1}{a}$, $b\rightarrow\frac{1}{b}$ and $c\rightarrow\frac{1}{c}$.
Hence, the condition it's the same and we need to prove that $\sum\limits_{cyc}\frac{ab}{2a^2+b}\leq1$.
But by C-S $$\sum\limits_{cyc}\frac{ab}{2a^2+b}=a+b+c+\sum\limits_{cyc}\left(\frac{ab}{2a^2+b}-a\right)=$$ $$=a+b+c-2\sum\limits_{cyc}\frac{a^3}{2a^2+b}=a+b+c-2\sum\limits_{cyc}\frac{a^4}{2a^3+ab}\leq$$ $$\leq a+b+c-\frac{2(a^2+b^2+c^2)^2}{2(a^3+b^3+c^3)+ab+ac+bc}$$ and it remains to prove that $$a+b+c-\frac{2(a^2+b^2+c^2)^2}{2(a^3+b^3+c^3)+ab+ac+bc}\leq1,$$ which is wrong for $b=c$.
Thank you!
The BW helps.
Let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.
Hence, we need to prove that $$\sum_{cyc}\frac{a}{a^2+\frac{2b(ab+ac+bc)}{a+b+c}}\leq\frac{a+b+c}{ab+ac+bc}$$ or $$\sum_{cyc}\frac{a}{a^2(ab+ac+bc)+2b(a+b+c)}\leq\frac{1}{ab+ac+bc}$$ or $$63(u^2-uv+v^2)a^7+(149u^3-11u^2v+105uv^2+149v^3)a^6+$$ $$+(145u^4-59u^3v-30u^2v^2+373uv^3+145v^4)a^5+$$ $$+(67u^5+33u^4v-164u^3v^2+268u^2v^3+357uv^4+67v^5)a^4+$$ $$+(12u^6+42u^5v-88u^4v^2+7u^3v^3+280u^2v^4+154uv^5+12v^6)a^3+$$ $$+2(5u^5-3u^4v-19u^3v^2+33u^2v^3+57uv^4+13v^5)uva^2+$$ $$+(2u^4-5u^3v-2u^2v^2+27uv^3+18v^4)u^2v^2a+4u^3v^6\geq0,$$ which is obvious.