Question

demonstrati inegalitatea$\sqrt{ \frac{x(y+z)}{3} } + \sqrt{ \frac{y(x+z)}{3} } + \sqrt{ \frac{z(x+y)}{3} } \leq \frac{5}{6} (x+y+z)$ , a,b,c apartin R

Answer 1

$\displaystyle \sqrt{ \frac{x(y+z)}{3} } = \sqrt{x \cdot \frac{y+z}{3} } \leq \frac{x+ \frac{y+z}{3} }{2} = \frac{3x+y+z}{6}. \\ \\ Scriem~si~relatiile~analoage: \\ \\ \sqrt{ \frac{y(x+z)}{3} } \leq \frac{3y+x+z}{6}~;~ \sqrt{ \frac{z(x+y)}{3} } \leq \frac{3z+x+y}{6}. \\ \\ Prin~insumarea~acestor~trei~relatii,~se~obtine~concluzia. \\ \\ Egalitatea~are~loc~cand~x= \frac{y+z}{3}~;~y= \frac{x+z}{3}~;~ z=\frac{x+y}{3}. \Leftrightarrow \\ \\ \Leftrightarrow 3x=y+z~;~3y=x+z~;~3z=x+y \Rightarrow$

$\displaystyle \Rightarrow 3(x+y+z)=2(x+y+z) \Rightarrow x+y+z=0 \Rightarrow y+z=-x~si \\ \\ relatiile~analoage.~Insa~y+z=3x~(si~relatiile~analoage) \RIghtarrow~ 3x=-x, \\ \\ 3y=-y,~3z=-z \Rightarrow egalitate ~pentru ~x=y=z=0. \\ \\ ^*partea~cu~"egalitatea~are~loc~cand"~este~optionala...$