Let $x,y,z$ be positive real numbers. We have to prove that:
$3(x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)\geq (x+y+z)^2(xy+yz+zx)^2.$
Thanks :)
Let $x,y,z$ be positive real numbers. We have to prove that:
$3(x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)\geq (x+y+z)^2(xy+yz+zx)^2.$
Thanks :)
Note that $x^2+xy+y^2=\frac{3}{4}(x+y)^2+\frac{1}{4}(x-y)^2\ge \frac{3}{4}(x+y)^2 $
Repeating the argument for y and z and then for x and z,we get $3(x^2+y^2+z^2)(y^2+yz+z^2)(x^2+xz+z^2)\ge \frac{81}{64}(x+y)^2(y+z)^2(z+x)^2$
or $8(x+y+z)(xy+yz+zx)\leq 9(x+y)(y+z)(z+x)\dots (1)$ iff $\sum x(y-z)^2\ge 0$ which is true! where the sum is taken cyclically over x,y and z.
P.S. Alternatively,you can proceed from (1) as follows:
$(x+y)(y+z)(z+x)=(x+y+z)(xy+xz+yz)-xyz$ which reduces the problem to proving $(x+y)(y+z)(z+x)\ge 8xyz$ which is true by AM-GM.
It is Indian NMO 2007, Problem 6. See here for lots of different solutions.