<s>[INST] If $x,y,z>0$, prove that $\frac{x+y+z}{3\sqrt{3}} \geq \frac{yz+zx+xy}{\sum \limits_{cyc} \sqrt{x^2+xy+y^2}}$ 
If $x,y,z > 0$, prove that $\dfrac{x+y+z}{3\sqrt{3}} \geq \dfrac{yz+zx+xy}{ \sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}}$
with equality if and only if $x=y=z$.
SOURCE :CRUX (Page Number 20 ; Question Number 805)

I tried various approaches like C-S, Holder and Schur, but failed. The problem is very cleverly devised because of which it seems invulnerable to the common tricks.
Any help will be gratefully acknowledged.
Thanks in Advance. :-)
 [/INST] Let $x^2+y^2+z^2=k(xy+xz+yz)$. 
Hence, $k\geq1$ and we need to prove that
$$\sum_{cyc}\sqrt{x^2+xy+y^2}\geq\frac{3\sqrt3(xy+xz+yz)}{x+y+z}$$ or
$$\sum_{cyc}(2x^2+xy)+2\sum_{cyc}\sqrt{(x^2+xy+y^2)(x^2+xz+z^2)}\geq\frac{27(xy+xz+yz)^2}{(x+y+z)^2}$$
Now, by C-S $$\sqrt{(x^2+xy+y^2)(x^2+xz+z^2)}\geq x^2+\frac{x(y+z)}{2}+yz.$$
Thus, it remains to prove that
$$\sum_{cyc}\left(2x^2+xy+2\left(x^2+\frac{x(y+z)}{2}+yz\right)\right)\geq\frac{27(xy+xz+yz)^2}{(x+y+z)^2}$$ or
$$\sum_{cyc}(4x^2+5xy)(x+y+z)^2\geq27(xy+xz+yz)^2$$ or
$$(4k+5)(k+2)\geq27,$$
which is obvious.
Done!
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