Dear Nikos,

By Nesbitt inequality, we get that U+V+W >= 3/2, thus by Chebyshev inequality we have

U(bc)^2 + V(ca)^2 + W(ab)^2

>= (1/3)(U+V+W)((bc)^2 + (ca)^2 + (ab)^2)

>= (1/2)((bc)^2 + (ca)^2 + (ab)^2),

hence is enough to prove that

(bc)^2 + (ca)^2 + (ab)^2 >= 16 F^2,

but (bc)^2 + (ca)^2 + (ab)^2 - 16 F^2 equals to

(1/2) ((b^2-c^2)^2 + (c^2-a^2)^2 + (a^2-b^2)^2),

which is always positive.

--- In Hyacinthos@yahoogroups.com, Nikolaos Dergiades <ndergiades@...> wrote:

>

> Dear friends,

> does anybody can give a simple proof to the

> following inequality?

> If p, q, r are arbitrary positive real numbers

> a,b,c the sides, F the area of triangle ABC

> and U = p/(q+r), V = q/(r+p), W = r/(p+q)

> then prove that

> U(bc)^2 + V(ca)^2 + W(ab)^2 >= 8F^2.

>

> Best regards

> Nikos Dergiades

>