Question

Suppose z>y>x>0 and $S=\frac{x^2+y^2+z^2}{x+y+z},then$

• A. $S>\frac{x^2}{z}$
• B. $S<\frac{z^2}{x}$
• C. $S,\frac{y^2}{x}$
• D. none of these

Answer 1

Answer: (A), (B)

$S<\frac{z^2}{x}$

Since z>y>x>0, we have $z^2>y^2>x^2.$
Therefore,
$3x^2<x^2+y^2+z^2<3z^2\left(1\right)and3x<x+y+z<3z\left(2\right)From\left(2\right),weget\frac{1}{3z}<\frac{1}{x+y+z}<\frac{1}{3x}\therefore \frac{3x^2}{3z}<\frac{x^2+y^2+z^2}{x+y+z}<\frac{3z^2}{3x}\Rightarrow \frac{x^2}{z}<\frac{x^2+y^2+z^2}{x+y+z}<\frac{z^2}{x}$