<s>[INST] Proving ${\frac {35{x}^{2}+7x(y+z)+23yz}{35(x^2+y^2+z^2)+37(xy+yz+zx)}}\leqslant \sqrt {{\frac {{x}^{2}+yz}{6\,{y}^{2} +6\,yz+6\,{z}^{2}}}}$ For $x,y,z \geqslant 0.$ Proving$:$
$${\dfrac {35{x}^{2}+7x(y+z)+23yz}{35(x^2+y^2+z^2)+37(xy+yz+zx)}}\leqslant \sqrt {{\dfrac {{x}^{2}+yz}{6{y}^{2}
+6yz+6{z}^{2}}}}\quad \quad(\text{tthnew})$$
My proof is using Buffalo Way$.$ By assuming $y\geqslant z,$ we only need to consider three cases:
$+)$ $x\geqslant y\geqslant z,$ let $x=z+u+v,y=z+u$ with helping by computer to simplify expression and we are done.
$+)$ $y\geqslant x \geqslant z,$ let $y=z+u+v,x=z+u$ and done after simplify..
$+)$ $y\geqslant z \geqslant x,$ we can prove by the similar way too!
And we are done! But it's not the nice proof. So any one have other$?$
Note. This inequality is a proof for Vo Quoc Ba Can's inequality$:$
$$\sqrt{\dfrac{a^2+bc}{b^2+bc+c^2}}+\sqrt{\dfrac{b^2+ca}{c^2+ca+a^2}}+\sqrt{\dfrac{c^2+ab}{a^2+ab+b^2}} \geqslant \sqrt{6}\,\,(\text{1})$$
$(\text{1})$ is inspired from the familiar Vasile Cîrtoaje's inequality$:$
$$\dfrac{a^2+bc}{b^2+bc+c^2}+\dfrac{b^2+ca}{c^2+ca+a^2}+\dfrac{c^2+ab}{a^2+ab+b^2} \geqslant 2$$
Also we have the following inequality by Crazy_LittleBoy (AoPS)
$${\frac { \left( {\frac {49\,\sqrt {2}}{12}}-{\frac {17}{3}} \right) x
\left( y+z \right) +{x}^{2}+ \left( \frac73-\frac76\sqrt {2} \right) yz}{{x
}^{2}+{y}^{2}+{z}^{2}+ \left( 7\,\sqrt {2}-9 \right) \left( xy+xz+yz
\right) }}\leq \sqrt {{\frac {{x}^{2}+yz}{6\,{y}^{2}+6\,yz+6\,{z}^{2}
}}}
$$
 [/INST] 
By Maple we have $$\sqrt{\frac{6(a^2+bc)}{b^2+bc+c^2}}\ge \frac{6a^2+4bc+ab+ac}{a^2+b^2+c^2+ab+bc+ca}$$
Note that $$\sum_{cyc} \frac{6a^2+4bc+ab+ac}{a^2+b^2+c^2+ab+bc+ca}=6$$
then we are done
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