Problem
ALG-B1-M07-P015 Sum of squares around the mean
#15
★★★☆☆ Level 3 of 5
Let \(a,b,c\) be real numbers and \(a+b+c=0\). Prove that \(a^2+b^2+c^2\ge0\), with equality only when \(a=b=c=0\). Then explain why this implies \(x^2+y^2+z^2\ge\frac{(x+y+z)^2}{3}\).
In the second part, subtract the average from \(x,y,z\).
The first part is obvious because a sum of squares is nonnegative; equality is possible only when all squares are zero.
For the second part, set \(m=\frac{x+y+z}{3}\), \(a=x-m\), \(b=y-m\), \(c=z-m\). Then \(a+b+c=0\), so \((x-m)^2+(y-m)^2+(z-m)^2\ge0\). Expanding gives \(x^2+y^2+z^2\ge3m^2=\frac{(x+y+z)^2}{3}\).
This is an important bridge to the arithmetic mean and quadratic mean.