My solution it is equialent to $\frac{bc}{2a^2+bc}+\frac{ab}{2c^2+ab}+\frac{ac}{2b^2+ac}\geq 1$ multiplying each side by $(2a^2+bc)(2b^2+ca)(2c^2+ab)$ it will be equialent to $a^4bc+ab^4c+abc^4\geq 3a^2b^2c^2$ which is true by A.M-G.M

$\frac{bc}{2a^2+bc}+\frac{ab}{2c^2+ab}+\frac{ac}{2b^2+ac}=\sum{\frac{(bc)^2}{2a^2bc+(bc)^2}}\geq \frac{(ab+bc+ca)^2}{2abc(a+b+c)+(ab)^2+(bc)^2+(ca)^2}=1$

Murad.Aghazade wrote: $\boxed{\text{A2}}$ Prove that for all Positive reals $a,b,c$ $\frac{a^2-bc}{2a^2+bc}+\frac{b^2-ca}{2b^2+ca}+\frac{c^2-ab}{2c^2+ab}\leq 0$ We have \[LHS=-\frac{3}{2}\frac{abc(a+b+c)\sum{(a-b)^2}}{(2a^2+bc)(ac+2b^2)(ab+2c^2)}\le{0}\]

Murad.Aghazade wrote: $\boxed{\text{A2}}$ Prove that for all Positive reals $a,b,c$ $\frac{a^2-bc}{2a^2+bc}+\frac{b^2-ca}{2b^2+ca}+\frac{c^2-ab}{2c^2+ab}\leq 0$ The inequality is equivalent to Let $x,y,z$ be Positive realssuch that $xyz=1$ .Prove that$$\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}\leq 1$$

I didnt understood how we got $\frac{bc}{2a^2+bc}+\frac{ab}{2c^2+ab}+\frac{ac}{2b^2+ac}\geq 1$ from $\frac{a^2-bc}{2a^2+bc}+\frac{b^2-ca}{2b^2+ca}+\frac{c^2-ab}{2c^2+ab}\leq 0$

sttsmet wrote: I didnt understood how we got $\frac{bc}{2a^2+bc}+\frac{ab}{2c^2+ab}+\frac{ac}{2b^2+ac}\geq 1$ from $\frac{a^2-bc}{2a^2+bc}+\frac{b^2-ca}{2b^2+ca}+\frac{c^2-ab}{2c^2+ab}\leq 0$ $\iff \frac{2(a^2-bc)}{2a^2+bc}+\frac{2(b^2-ca)}{2b^2+ca}+\frac{2(c^2-ab)}{2c^2+ab}\leq 0$ $\iff \frac{2a^2+bc-3bc}{2a^2+bc}+\frac{2b^2+ca-3ca}{2b^2+ca}+\frac{2c^2+ab-3ab}{2c^2+ab}\leq 0$ $ \iff 3- 3\left(\frac{bc}{2a^2+bc}+\frac{ab}{2c^2+ab}+\frac{ac}{2b^2+ac}\right )\geq 0 $

the first line miss a(

Mathskidd wrote: sttsmet wrote: I didnt understood how we got $\frac{bc}{2a^2+bc}+\frac{ab}{2c^2+ab}+\frac{ac}{2b^2+ac}\geq 1$ from $\frac{a^2-bc}{2a^2+bc}+\frac{b^2-ca}{2b^2+ca}+\frac{c^2-ab}{2c^2+ab}\leq 0$ $\iff \frac{2(a^2-bc)}{2a^2+bc}+\frac{2(b^2-ca)}{2b^2+ca}+\frac{2c^2-ab)}{2c^2+ab}\leq 0$ $\iff \frac{2a^2+bc-3bc}{2a^2+bc}+\frac{2b^2+ca-3ca}{2b^2+ca}+\frac{2c^2+ab-3ab}{2c^2+ab}\leq 0$ $ \iff 3- 3\left(\frac{bc}{2a^2+bc}+\frac{ab}{2c^2+ab}+\frac{ac}{2b^2+ac}\right )\geq 0 $ the first line miss a(

Kingbun-Kitty wrote: the first line miss a( Fixed

sqing wrote: Murad.Aghazade wrote: $\boxed{\text{A2}}$ Prove that for all Positive reals $a,b,c$ $\frac{a^2-bc}{2a^2+bc}+\frac{b^2-ca}{2b^2+ca}+\frac{c^2-ab}{2c^2+ab}\leq 0$ The inequality is equivalent to Let $x,y,z$ be Positive realssuch that $xyz=1$ .Prove that$$\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}\leq 1$$ $\frac{a^2-bc}{2a^2+bc}+\frac{b^2-ca}{2b^2+ca}+\frac{c^2-ab}{2c^2+ab}\leq 0$ $\iff \frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}\leq \frac{bc}{2a^2+bc}+\frac{ca}{2b^2+ca}+\frac{ab}{2c^2+ab}$ $\iff \frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}\leq \frac{2a^2+bc-2a^2}{2a^2+bc}+\frac{2b^2+ca-2b^2}{2b^2+ca}+\frac{2c^2+ab-2c^2}{2c^2+ab}$ $\iff 3\left(\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}\right)\leq 3$ Let $x=\frac{bc}{a^2}, y=\frac{ca}{b^2},z= \frac{ab}{c^2}$ and $xyz=1$ $$\iff \frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}\leq 1$$

IstekOlympiadTeam wrote: $\frac{bc}{2a^2+bc}+\frac{ab}{2c^2+ab}+\frac{ac}{2b^2+ac}=\sum{\frac{(bc)^2}{2a^2bc+(bc)^2}}\geq \frac{(ab+bc+ca)^2}{2abc(a+b+c)+(ab)^2+(bc)^2+(ca)^2}=1$ Titu's lemma

wrong solution

My solution: Multiply with -1, and then add 3 to both of the sides, 1 to each fraction (such that the expressions are positive) and then you just have to use AM-GM and you are done.

Multiply both sides by 2, and note $\frac{2a^2-2bc}{2a^2+bc} = 1-\frac{3bc}{2a^2+bc}$ Hence the inequality is equivalent to $\sum_{cyc} \frac{bc}{2a^2+bc} \geq 1$ Now let $x=bc, y=ac, z=ab$, so this becomes $\sum_{cyc} \frac{x^2}{2yz + x^2} \geq 1$. But as $2yz \leq y^2 + z^2$, $\sum_{cyc} \frac{x^2}{2yz + x^2} \geq \sum_{cyc} \frac{x^2}{x^2+y^2+z^2} = 1$.