In APMO, I solved this problem by following way. $\sum_{cyc}\frac{x^{2}+yz}{\sqrt{2x^{2}(y+z)}}=\sum_{cyc}\frac{(x-y)(x-z)}{\sqrt{2x^{2}(y+z)}}+\sum_{cyc}\sqrt{\frac{y+z}{2}}$. But $\sum_{cyc}\sqrt{\frac{y+z}{2}}\ge \sum_{cyc}\frac{\sqrt{y}+\sqrt{z}}{2}=1$, WTS : $\sum_{cyc}\frac{(x-y)(x-z)}{\sqrt{2x^{2}(y+z)}}\ge0$ WLOG $x\ge y\ge z$ -> $\frac{1}{y\sqrt{z+x}}\le \frac{1}{z\sqrt{x+y}}$. Done.

$x,y,z$ are positive reals such that $\sqrt{x}+\sqrt{y}+\sqrt{z}=1$ then \[sum_{\substack{cyc}}\frac{x^{2}+yz}{\sqrt{2x^{2}(y+z)}}=\sum_{\substack{cyc}}\frac{x}{\sqrt{2(y+z)}}+\sum_{\substack{cyc}}\frac{yz}{x\sqrt{2(y+z)}}\] now we have : lemma 1:$\sum_{\substack{}}\frac{x}{\sqrt{2(y+z)}}\ge\frac{1}{2}.$ proof:we apply jensen to get above cyclic sum as \[\ge \sqrt{\frac{(x+z+y)^{3}}{4(xy+zx+zy)}}\ge \frac{\sqrt{3}}{2}.\sqrt{x+y+z}\ge 1/2 \] \[as\ x+y+z \ge 1/3.(\sqrt{x}+\sqrt{y}+\sqrt{z})^{2}=1/3 \ and\ (x+z+y)^{2}\ge 3(xz+zy+xy) \] now lemma 2:$\sum_{\substack{sym}}\frac{yz}{x\sqrt{2(y+z)}}\ge 1/2$ proof : again applying jensen we have above cyclic expression as \[\ge \frac{1}{xyz}.\sqrt{\frac{(x^{2}z^{2}+x^{2}y^{2}+y^{2}z^{2})^{3}}{2.\sum_{\substack{sym}}x^{3}y^{2}}}\ge \frac{\sqrt{3(x+z+y)}}{2}\] $iff \sum_{\substack{sym}}x^{6}y^{6}+6\sum_{\substack{sym}}x^{6}y^{4}z^{2}+12x^{4}y^{4}z^{4}\ge 3\sum_{\substack{sym}}x^{6}y^{4}z^{2}+3\sum_{\substack{sym}}x^{5}y^{5}z^{2}+3\sum_{\substack{sym}}x^{5}y^{4}z^{3}$ from schur inequality \[\sum_{\substack{sym}}x^{6}y^{6}+6x^{4}y^{4}z^{4}\ge 2\sum_{\substack{sym}}x^{6}y^{4}z^{2}\] so we have to prove $3(\sum_{\substack{sym}}x^{6}y^{4}z^{2}-x^{5}y^{5}z^{2})+2(\sum_{\substack{sym}}x^{6}y^{4}z^{2}+1/2(x^{4}z^{4}y^{4}))-3\sum_{\substack{sym}}x^{5}y^{4}z^{3}\ge 0$by AM-GM and MUIRHEAD inequality adding two lemmas we have the result.

i made a different solution... it's key is... faith from the condition and AM-GM we have that $\sum_{cyc}\frac{x^{2}+yz}{\sqrt{2x^{2}(y+z)}}=\sum_{cyc}\frac{x^{2}+yz}{x(\sqrt{2x(y+z)}+\sqrt{2y(y+z)}+\sqrt{2z(y+z)})}\geq \sum_{cyc}\frac{2(x^{2}+yz)}{x(2x+5y+5z)}$... the rest is plain muirhead. after clearing the denominators, it's exagerated how so much terms cancell and the thing to prove at last falls so easy from muirhead...

Note that by Chebyshev, $\sum \frac{2yz-xy-xz}{\sqrt{2x^{2}(y+z)}}\ge 0$ $\sum \frac{2x^{2}+2yz-2x^{2}-xy-xz}{\sqrt{2x^{2}(y+z)}}\ge 0$ $\sum \frac{2x^{2}+2yz-2\sqrt{2x^{3}(y+z)}}{\sqrt{2x^{2}(y+z)}}\ge 0$ $\sum \frac{x^{2}+yz}{\sqrt{2x^{2}(y+z)}}\ge \sum \sqrt{x}= 1$.

Here is my solution Letr $x=a^{2},y=b^{2},z=c^{2}$ $\implies$ $a+b+c=1$ $\sum\frac{a^{4}+b^{2}c^{2}}{a\sqrt{2a^{2}(b^{2}+c^{2})}}\geq 1$ from Cauchy $\sum\frac{a^{4}+b^{2}c^{2}}{a\sqrt{2a^{2}(b^{2}+c^{2})}}\sum{a^{2}\sqrt{2a^{2}(b^{2}+c^{2})}}\geq (\sum\sqrt{a^{4}+b^{2}c^{2}})^{2}$ $a^4+b^4+c^4+a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2}+ 2\sum \sqrt{(a^{4}+b^{2}c^{2})(b^{4}+c^{2}a^{2})}\geq \sum{a\sqrt{2a^{2}(b^{2}+c^{2})}}$ from AM-GM $\sqrt{2a^{2}(b^{2}+c^{2})}\leq a^{2}+\frac{b^{2}+c^{2}}{2}$ $\implies$ $\sum{a\sqrt{2a^{2}(b^{2}+c^{2})}}\leq (\sum a^{3})+ \frac{[2,1,0]}{2}=\frac{[3,0,0]+[2,1,0]}{2}$ $\frac{[3,0,0]+[2,1,0]}{2}=\frac{[3,0,0][1,0,0]+[2,1,0][1,0,0]}{4}=\frac{[4,0,0]}{2}+ \frac{3[3,1,0]}{2}+\frac{[2,1,1]+[2,2,0]}{2}$ $2\sum \sqrt{(a^{4}+b^{2}c^{2})(b^{4}+c^{2}a^{2})} \geq 2[3,1,0]$ (Cauchy) $a^4+b^4+c^4+a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2}+ 2\sum \sqrt{(a^{4}+b^{2}c^{2})(b^{4}+c^{2}a^{2})}\geq 2[3,1,0]+\frac{[4,0,0]+[2,2,0]}{2}$ $2[3,1,0]+\frac{[4,0,0]+[2,2,0]}{2}\geq\frac{[4,0,0]}{2}+ \frac{3[3,1,0]}{2}+\frac{[2,1,1]+[2,2,0]}{2}$ this is equivalent to $[3,1,0]\geq [2,1,1]$ which is obvious Q.E.D

We have \[ \begin {align*} \sum \frac {x^2 + yz} {\sqrt {2x^2 (y + z)}} = \sum \frac {x^2 + yz} {x(y + z)} \cdot \sqrt {\frac {y + z} {2}} \ge \sum \frac {x^2 + yz} {x(y + z)} \left (\frac {\sqrt y + \sqrt z} {2} \right) \end {align*} \] Now \[ \begin {align*} \sum \frac {x^2 + yz} {x(y + z)} \left (\frac {\sqrt y + \sqrt z} {2} \right) &\ge 1 \iff\\ \sum \frac {(x + y) (x + z)} {x (y + z)} \left (\frac {\sqrt y + \sqrt z} {2}\right) & \ge 2 \iff \\ \sum \frac {\sqrt y} {2} \left (\frac {(x + y)(x + z)} {x(y + z)} + \frac {(z + y) (z + x)} {z (x + y)}\right) & \ge 2 \end {align*} \] From AM-GM, \[ \begin {align*} \frac {\sqrt y} {2} \left (\frac {(x + y)(x + z)} {x(y + z)} + \frac {(z + y) (z + x)} {z (x + y)}\right) \ge (x + z) \sqrt \frac {y} {xz} \ge 2 \sqrt y \end {align*} \] Summing cyclically gives us the desired result.

I did same as Little Gauss

I have a generalization for this in cases of : $$n=3$$

Gen1 was here