For $x,\,y,\,z> 0$, prove that $$\frac{y}{\sqrt{2\,z(\,x+ y\,)}}+ \frac{z}{\sqrt{2\,x(\,y+ z\,)}}+ \frac{x}{\sqrt{2\,y(\,z+ x\,)}}\geqq \frac{3}{2}$$
I tried $\lceil$ HOLDER!inequality $\rfloor$ with integer polynomial but unsucessful , so I used else methods!
Such as
By a.m.-g.m.-inequality, we have $$\sqrt{2\,z(\,x+ y\,)}\leqq \frac{2\,z+ x+ y}{2}$$ Thus, we need to prove $$\frac{2\,y}{2\,z+ x+ y}+ \frac{2\,z}{2\,x+ y+ z}+ \frac{2\,x}{2\,y+ z+ x}\geqq \frac{3}{2}$$ By Titu-lemma $$\frac{(\,y+ z+ x\,)^{\,2}}{y(\,2\,z+ x+ y\,)+ z(\,2\,x+ y+ z\,)+ x(\,2\,y+ z+ x\,)}\geqq \frac{3}{4}$$ is true, but I can't use $\lceil$ HOLDER!inequality $\rfloor$ here! I need to the help, I used these to prep for my book, thanks!
By Holder $$\sum_{cyc}\frac{x}{\sqrt{y(x+z)}}=\sqrt{\frac{\left(\sum\limits_{cyc}\frac{x}{\sqrt{y(x+z)}}\right)^2\sum\limits_{cyc}xy(x+z)}{\sum\limits_{cyc}xy(x+z)}}\geq\sqrt{\frac{(x+y+z)^3}{\sum\limits_{cyc}xy(x+z)}}.$$ Thus, it's enough to prove that $$2(x+y+z)^3\geq9\sum\limits_{cyc}xy(x+z)$$ or $$\sum_{cyc}(2x^3-3x^2y+6x^2z-5xyz)\geq0,$$ which is true by AM-GM and SOS: $$\sum_{cyc}(2x^3-3x^2y+6x^2z-5xyz)\geq\sum_{cyc}(2x^3-3x^2y+xy^2)=$$ $$=\sum_{cyc}x(x-y)(2x-y)=\sum_{cyc}\left((x-y)(2x^2-xy)-\frac{x^3-y^3}{3}\right)=$$ $$=\frac{1}{3}\sum_{cyc}(x-y)^2(5x+y)\geq0.$$