Problem
ALG-B2-M07-P003 Sum normalization
#3
★★☆☆☆ Level 2 of 5
Prove for \(a,b,c\ge0\): \[a^2+b^2+c^2\ge\frac{(a+b+c)^2}{3}.\]
Hint. Set \(a+b+c=1\).
The inequality is homogeneous. With \(a+b+c=1\), we need \(a^2+b^2+c^2\ge1/3\). This follows from \((a+b+c)^2\le3(a^2+b^2+c^2)\). Returning to the original scale gives the result.
Basic use of sum normalization.