Find the largest real $ T$ such that for each non-negative real numbers $ a,b,c,d,e$ such that $ a+b=c+d+e$: \[ \sqrt{a^{2}+b^{2}+c^{2}+d^{2}+e^{2}}\geq T(\sqrt a+\sqrt b+\sqrt c+\sqrt d+\sqrt e)^{2}\]
Problem
Source: Iranian National Olympiad (3rd Round) 2007
Tags: inequalities, inequalities proposed
the_lone_wolf
02.09.2007 03:35
I think T = 1/ sqrt(125) . Is it correct?
Albanian Eagle
04.09.2007 19:23
by putting $ a=b=3$ and $ c=d=e=2$ we find $ \frac{\sqrt{30}}{6(\sqrt{3}+\sqrt{2})^{2}}\geq T$ to prove that for such a $ T$ the inequality always holds notice that $ x^{2}+y^{2}\geq 2(\frac{x+y}{2})^{2}$ and $ x^{2}+y^{2}+z^{2}\geq 3(\frac{x+y+z}{3})^{2}$ which means that the worst case is when $ a=b$ and $ c=d=e$ for which we have equality
Stranger8
05.04.2016 07:54
My solution is the same as Zarif