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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. :-)

  • 0
    Why the Down-vote? – 2017-02-19
  • 0
    Geometrically, this problem says that, in a triangle $ABC$ all of whose angles are less than or equal to $\frac{2\pi}{3}$, the sum $PA+PB+PC$ is greater than or equal to $6r$ for any point $P$ on the plane. Here, $r$ is the inradius of $ABC$. It would be great to see a geometric proof. – 2017-08-24

1 Answers 1

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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!