19889Re: Geometric inequality
- Mar 5, 2011Dear 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)hence is enough to prove that
>= (1/2)((bc)^2 + (ca)^2 + (ab)^2),
(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
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