Prove $\frac{x^2+yz}{\sqrt{2x^2(y+z)}}+\frac{y^2+zx}{\sqrt{2y^2(z+x)}}+\frac{z^2+xy}{\sqrt{2z^2(x+y)}}\geqq 1$ For $x,y,z>0$ and $\sqrt{x} +\sqrt{y} +\sqrt{z} =1.$ Prove that$:$
$$\frac{x^2+yz}{\sqrt{2x^2(y+z)}}+\frac{y^2+zx}{\sqrt{2y^2(z+x)}}+\frac{z^2+xy}{\sqrt{2z^2(x+y)}}\geq 1$$
My solution$:$
Let $x=a^2,\,y=b^2,\,z=c^2$ then $a+b+c=1,$ we need to prove$:$ $$\sum\limits_{cyc} \frac{a^4+b^2 c^2}{a^2 \sqrt{2(b^2+c^2)}} \geqq 1\Leftrightarrow  \sum\limits_{cyc} \frac{a^4+b^2 c^2}{a^2 \sqrt{2(b^2+c^2)}} \geqq a+b+c$$
By AM-GM$:$ $$\text{LHS} = \sum\limits_{cyc} \frac{a^4+b^2 c^2}{a \sqrt{2a^2(b^2+c^2)}} \geqq \sum\limits_{cyc} \frac{2(a^4+b^2c^2)}{a(2a^2+b^2+c^2)} \geqq a+b+c$$
Last inequality is true by SOS$:$
$$\sum\limits_{cyc} \frac{2(a^4+b^2c^2)}{a(2a^2+b^2+c^2)}-a-b-c=\sum\limits_{cyc} {\frac {{c}^{2} \left( a-b \right) ^{2} \left( a+b \right)  \left( 2\, {a}^{2}+ab+2\,{b}^{2}+{c}^{2} \right) }{a \left( 2\,{a}^{2}+{b}^{2}+{c }^{2} \right) b \left( {a}^{2}+2\,{b}^{2}+{c}^{2} \right) }} \geqq 0$$
PS:Is there any another solution for original inequality or the last inequality$?$
Thank you!

Another way.
Since $$\left(yz-\frac{1}{2}xy-\frac{1}{2}zx, zx-\frac{1}{2}yz-\frac{1}{2}xy,xy-\frac{1}{2}zx-\frac{1}{2}yz\right)$$ and
$$\left(\frac{1}{x\sqrt{2(y+z)}},\frac{1}{y\sqrt{2(z+x)}},\frac{1}{z\sqrt{2(x+y)}}\right)$$ have the same ordering, by AM-GM and Chebyshov we obtain:
$$\sum_{cyc}\frac{x^2+yz}{x\sqrt{2(y+z)}}-1=\sum_{cyc}\left(\frac{x^2+yz}{x\sqrt{2(y+z)}}-\sqrt{x}\right)=$$
$$=\sum_{cyc}\frac{x^2+yz-x\sqrt{2x(y+z)}}{x\sqrt{2(y+z)}}\geq \sum_{cyc}\frac{x^2+yz-\frac{1}{2}x(2x+y+z)}{x\sqrt{2(y+z)}}=$$
$$=\sum_{cyc}\frac{yz-\frac{1}{2}xy-\frac{1}{2}zx}{x\sqrt{2(y+z)}}\geq\frac{1}{3}\sum_{cyc}\left(yz-\frac{1}{2}xy-\frac{1}{2}zx\right)\sum_{cyc}\frac{1}{x\sqrt{2(y+z)}}=0.$$
