$\frac {3 x y z}{(x+y) (y+z) (z+x)} + \sum\limits_{cycl}^{} \left(\frac {x+y}{x+y+ 2 z}\right)^2 \ge \frac {9}{8}.$

For $x,y,z>0,$ I have to prove that $$\frac {3 x y z}{(x+y) (y+z) (z+x)} + \sum\limits_{cycl}^{} \left(\frac {x+y}{x+y+ 2 z}\right)^2 \ge \frac {9}{8}.$$

I tried to use $$\sqrt {\frac {1}{3} \sum\limits_{cycl}^{} (\frac {x+y}{x+y+ 2 z})^2} \ge \frac {1}{3} \sum\limits_{cycl}^{} \frac {x+y}{x+y+ 2 z} \ge \sqrt[3] {(\frac {x+y}{x+y+ 2 z}) (\frac {y+z}{2 x +y +z}) (\frac {z + x}{z+2 y+z})},$$

but it looks hard to continue and $\frac {3 x y z}{(x+y) (y+z) (z+x)}$ seems to be not helpful. Any help? Thank you.

• If you do not get an answer, I would strongly suggest posting to AOPS. Specifically, post in High School Olympiads. – greenturtle3141 Jul 18 '18 at 22:04
• it is not possible to solve it in AOPS, in High School Olympiads. Many of the questions are unanswered there – Steven Jul 18 '18 at 22:10
• I beg to differ. Many problems there get an answer. People trained in solving these types of math competition problems are quite often found in HSO. – greenturtle3141 Jul 18 '18 at 22:15
• is there any solution? – Steven Jul 19 '18 at 12:53

Let $\frac{2x}{y+z}=a$, $\frac{2y}{x+z}=b$ and $\frac{2z}{x+y}=c$.

Thus, $ab+ac+bc+abc=4$ and we need to prove that $$\sum_{cyc}\frac{1}{(a+1)^2}+\frac{3}{8}abc\geq\frac{9}{8}.$$ Now, let $a+b+c=3u$, $ab+ac+bc=3v^2,$ where $v>0$ and $abc=w^3$.

Hence, the condition does not depend on $u$.

In another hand, by AM-GM $$4=ab+ac+bc+w^3=3v^2+w^3\leq3v^2+v^3,$$ which gives $v\geq1$ and since $u\geq v$, we obtain $u\geq1$.

Also, $$4=3v^2+w^3\geq3w^2+w^3,$$ which gives $w\leq1$.

Now, we need to prove that $$\frac{\sum\limits_{cyc}(ab+a+b+1)^2}{(abc+ab+ac+bc+a+b+c+1)^2}+\frac{3}{8}w^3\geq0$$ or $$\frac{\sum\limits_{cyc}(a^2b^2+2a^2+1+2a^2b+2a^2c+4ab+4a)}{(w^3+3v^2+3u+1)^2}+\frac{3}{8}w^3\geq\frac{9}{8}$$ or $$\frac{9v^4-6uw^3+18u^2-12v^2+3+18uv^2-6w^3+12v^2+12u}{(w^3+3v^2+3u+1)^2}+\frac{3}{8}w^3\geq\frac{9}{8}$$ or $f(u)\geq0,$ where $$f(u)=\frac{3v^4-2uw^3+6u^2+1+6uv^2-2w^3+4u}{(w^3+3v^2+3u+1)^2}+\frac{1}{8}w^3-\frac{3}{8}.$$ But $$f'(u)=\tfrac{-2w^3+12u+6v^2+4}{(w^3+3v^2+3u+1)^2}-\tfrac{2\cdot3(3v^4-2uw^3+6u^2+1+6uv^2-2w^3+4u)}{(w^3+3v^2+3u+1)^3}=$$ $$=\frac{2(9uw^3+9uv^2+9v^2+7w^3-w^6-1)}{(w^3+3v^2+3u+1)^3}>0,$$ which says that $f$ increases and since the condition does not depend on $u$, it's enough to prove our inequality for a minimal value of $u$, which happens for an equality case of two variables.

Let $b=a$.

Hence, the condition gives $$a^2+2ac+a^2c=4$$ or $$ac(2+a)=4-a^2$$ or $$c=\frac{2-a}{a},$$ where $0<a<2$ and we need to prove that $$\frac{2}{(a+1)^2}+\frac{1}{\left(\frac{2-a}{a}+1\right)^2}+\frac{3}{8}a^2\cdot\frac{2-a}{a}\geq\frac{9}{8}$$ or $$(a-1)^2(7+2a-a^2)\geq0,$$ which is obvious.

Done!