Prove that $(x+y+z)^2(yz+zx+xy)^2 \leq 3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2)$ This is Problem 6 of the 2007 Indian National Math Olympiad (INMO).

If $x, y, z$ are positive real numbers, prove that
$(x+y+z)^2(yz+zx+xy)^2 \leq 3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2).$

My best idea was to expand this and simplify. Although that doesn't look very feasible. Another idea is to see that $x^2+y^2+xy \geq x^2+y^2$. Then we just have to show that $(x+y+z)^2(yz+zx+xy)^2 \leq 3(x^2+y^2)(x^2+z^2)(y^2+z^2)$ if that is even true.
 A: $(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2xz \\ \le x^2+xy+y^2+y^2+yz+z^2+x^2+xz+z^2$
$3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2)-(x^2+xy+y^2+y^2+yz+z^2+x^2+xz+z^2)(yz+zx+xy)^2=(x+y+z)^2(y^2z^2-xyz^2+x^2z^2-xy^2z-x^2yz+x^2y^2) \ge0$
A: 
Let $AF=x\ , BF=y\ , CF=z$. 
$F-$ Fermat point of $\triangle ABC\ $
So inequality we can rewrite as : $x+y+z\le 3R$ , which is obviously true)
A: Let I = 3(x² + xy + y²)(y² + yz + z²)(z² + zx + x²)
Then, By Holder's Inequality, it follows that
I >= 81x²y²z²
Thus, we need to prove that
81x²y²z² >= (x + y + z)²(xy + yz + zx)²
Or 9xyz >= (x + y + z)(xy + yz + zx)
Expanding R.H.S. gives
9xyz >= xy(x + y) + yz(y + z) + zx(z + x) + 3xyz
Or  xy(x + y) + yz(y + z) + zx(z + x) >= 6xyz
Which is evidently true by AM>=GM. QED
A: We'll prove that our inequality is true for all reals $x$, $y$ and $z$.
Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Since $\prod\limits_{cyc}(x^2+xy+y^2)=27(3u^2v^4-u^3w^3-v^6)$, 
we see that our inequality is equivalent to $f(w^3)\geq0$, where $f$ is a linear function.
Thus, $f$ get's a minimal value for an extremal value of $w^3$.
Since $x$, $y$ and $z$ are real roots of the equation $(X-x)(X-y)(X-z)=0$ or
$X^3-3uX^2+3v^2X-w^3=0$ or $X^3-3uX^2+3v^2X=w^3$, 
we see that a line $Y=w^3$ and a graph $Y=X^3-3uX^2+3v^2X$ should have three common points and the extremal value of $w^3$ happens for the equality case of two variables.
Since our inequality is homogeneous and even degree, it remains to check one case only: $y=z=1$, 
which gives $(x-1)^2(5x^2+8x+5)\geq0$. Done!
