Let $a = \frac{x}{y}$, $b = \frac{y}{z}$, $c = \frac{z}{x}$, and clear denominators, we have: $\sum \frac{y^2z^2}{x^2z^2+x^2y^2+xyz^2+xy^2z} \geq \frac{3}{4}$. Now let $p = y^2z^2$, $q = z^2x^2$, $r = x^2y^2$, so we have to prove that $\sum \frac{p}{q+r+\sqrt{pq}+\sqrt{pr}} \geq \frac{3}{4}$ We have $\sum \frac{p}{q+r+\sqrt{pq}+\sqrt{pr}} \geq \sum \frac{p}{q+r+\frac{p+q}{2}+\frac{p+r}{2}} = \sum\frac{2p}{2p+3q+3r}$ The inequality is homogeneous, so we can assume $p+q+r = 3$. $\sum\frac{2p}{2p+3q+3r} = \sum\frac{2p}{9-p}$ But for $x \in (0,3)$ we have $\frac{2x}{9-x} \geq \frac{9x-1}{32} \Leftrightarrow \frac{9(x-1)^2}{32(9-x)} \geq 0$ So $\sum\frac{2p}{9-p} \geq \sum\frac{9p-1}{32} = \frac{9(p+q+r)-3}{32} = \frac{3}{4}$, qed.

Let $a = \frac{x}{y}$, $b = \frac{y}{z}$, $c = \frac{z}{x}$, and clear denominators, we have: $\sum \frac{y^2z^2}{x^2z^2+x^2y^2+xyz^2+xy^2z} \geq \frac{3}{4}$. Now this is just Titu's lemma

I have got another solution, but I'm not sure if it is correct, Let denote $ S=\frac{1}{a(a+1)+ab(ab+1)}+\frac{1}{b(b+1)+bc(bc+1)}+\frac{1}{c(c+1)+ca(ca+1)} $ Then we will have to prove that, $ S\geq\frac{3}{4} $ $ S=\frac{1}{a(a+1)+ab(ab+1)}+\frac{1}{b(b+1)+bc(bc+1)}+\frac{1}{c(c+1)+ca(ca+1)}=\frac{abc}{a^2+a+(ab)^2+ab)}+\frac{abc}{b^2+b+(bc)^2+bc}+\frac{abc}{c^2+c+(ca)^2+ca}=\frac{bc}{a+1+ab^2+b}+\frac{ac}{b+1+bc^2+c}+\frac{ab}{c+1+ca^2+a}=\frac{bc}{a+1+\frac{b}{c}+b}+\frac{ac}{b+1+\frac{c}{a}+c}+\frac{ab}{c+1+\frac{a}{b}+a} $ So we have by Holder's inequality that, $ S[(a+1+\frac{b}{c}+b)+(b+1+\frac{c}{a}+c)+(c+1+\frac{a}{b}+a)][(ab)^2+(bc)^2+(ca)^2]\geq(ab+bc+ca)^3$ or $ S[2(a+b+c)+(\frac{b}{c}+\frac{c}{a}+\frac{a}{b})+3][(ab)^2+(bc)^2+(ca)^2]\geq(ab+bc+ca)^3 $ we multiply both sides with $36$ and we have, $ 36S[2(a+b+c)+(\frac{b}{c}+\frac{c}{a}+\frac{a}{b})+3][(ab)^2+(bc)^2+(ca)^2]\geq36(ab+bc+ca)^3 $ $(1)$ we know from the AM-GM inequality that, $[2(a+b+c)+(\frac{b}{c}+\frac{c}{a}+\frac{a}{b})+3][(ab)^2+(bc)^2+(ca)^2]\geq(6+3+3)3=36$ $(2)$ So from $(1)$ and $(2)$ we understand that $36S\geq(ab+bc+ca)^3$ But $(ab+bc+ca)^3\geq 27$ So $36S\geq 27 \implies S\geq \frac{27}{36} \implies S\geq \frac{3}{4}$ which is what we wanted to show!

nkalosidhs wrote: we multiply both sides with 36 and we have, $36S[2(a+b+c)+(\frac{b}{c}+\frac{c}{a}+\frac{a}{b})+3][(ab)^2+(bc)^2+(ca)^2]\geq36(ab+bc+ca)^3$ (1) we know from the AM-GM inequality that, $[2(a+b+c)+(\frac{b}{c}+\frac{c}{a}+\frac{a}{b})+3][(ab)^2+(bc)^2+(ca)^2]\geq(6+3+3)3=36$ (2) So from (1) and (2) we understand that $36S\geq(ab+bc+ca)^3$ This part is wrong,if $ab\ge cd$ and $a\ge c$,then we don't always have $b\ge c$,take for example $a=2,b=3,c=1,d=5$.

hurricane, I think you are right...I've mistaken in that part...thanks for the notice!

utkarshgupta wrote: Let $a = \frac{x}{y}$, $b = \frac{y}{z}$, $c = \frac{z}{x}$, and clear denominators, we have: $\sum \frac{y^2z^2}{x^2z^2+x^2y^2+xyz^2+xy^2z} \geq \frac{3}{4}$. Now this is just Titu's lemma Can u explain in details please ? Using Titu's lemma i got a stronger result.

What's wrong in getting a stronger result

utkarshgupta wrote: What's wrong in getting a stronger result $ \sum\frac{y^2z^2}{x^2z^2+x^2y^2+xyz^2+xy^2z}\geq\frac{(xy+yz+zx)^2}{2(\sum x^2y^2)+2xyz(x+y+z)}$ we now have to show $\frac{(xy+yz+zx)^2}{2(\sum x^2y^2)+2xyz(x+y+z)}\geq\frac{3}{4}$ $\implies xyz(x+y+z) \geq\sum x^2y^2$