Prove $\frac{x^2+yz}{\sqrt{2x^2(y+z)}}+\frac{y^2+zx}{\sqrt{2y^2(z+x)}}+\frac{z^2+xy}{\sqrt{2z^2(x+y)}}\geqq 1$

Let $x\geq y\geq z$.

Thus, by C-S we obtain: $$\sum_{cyc}\frac{x^2+yz}{\sqrt{2x^2(y+z)}}=\sum_{cyc}\frac{x^2-xy-xz+yz}{\sqrt{2x^2(y+z)}}+\sum_{cyc}\frac{xy+xz}{\sqrt{2x^2(y+z)}}=$$ $$=\sum_{cyc}\frac{(x-y)(x-z)}{\sqrt{2x^2(y+z)}}+\sum_{cyc}\frac{\sqrt{2(y+z)}}{2}\geq$$ $$\geq \frac{(x-y)(x-z)}{\sqrt{2x^2(y+z)}}+\frac{(y-x)(y-z)}{\sqrt{2y^2(x+z)}}+\frac{1}{2}\sum_{cyc}(\sqrt{y}+\sqrt{z})=$$ $$=\frac{x-y}{\sqrt2}\left(\frac{x-z}{x\sqrt{y+z}}-\frac{y-z}{y\sqrt{x+z}}\right)+1\geq$$ $$\geq \frac{x-y}{\sqrt2}\left(\frac{\frac{x}{y}(y-z)}{x\sqrt{y+z}}-\frac{y-z}{y\sqrt{x+z}}\right)+1=$$ $$=\frac{(x-y)(y-z)}{y\sqrt2}\left(\frac{1}{\sqrt{y+z}}-\frac{1}{\sqrt{x+z}}\right)+1\geq1.$$