Question

Determinati numerele reale x,y,z pentru care:
$\sqrt{x-2010} + \sqrt{y+2012} + \sqrt{z-4} = \frac{x+y+z+1}{2}$

Answer 1

$Simplu~ :P \\ \\ Din~inegalitatea~mediilor~(m_g \leq m_a)~avem: \\ \\ \sqrt{a}= \sqrt{a \cdot 1} \leq \frac{a+1}{2}~\forall~a \geq0~(egalitate~\Leftrightarrow a=1) \\ \\ Deci~\sqrt{x-2010}+ \sqrt{y+2012}+ \sqrt{z-4} \leq \frac{x-2010+1}{2}+ \frac{y+2012+1}{2}+ \\ \\ + \frac{z-4+1}{2}=\frac{x+y+z+1}{2}. \\ \\ Egalitatea~are~loc~daca~si~numai~daca~x-2010=1~;~ \\ \\ y+2012=1~si~z-4=1,~adica~pentru~ \\ \\ (x,y,z)=(2011,-2011,5).$