Problem
ALG-B2-M07-P014 Sum of squares with sum 3
#14
★★★★☆ Level 4 of 5
Let \(a,b,c\ge0\), \(a+b+c=3\). Prove \[a^2+b^2+c^2\ge3.\]
Hint. This is the normalized version of \(\sum a^2\ge\frac{(a+b+c)^2}{3}\).
By Cauchy, \((a+b+c)^2\le3(a^2+b^2+c^2)\). Since \(a+b+c=3\), we get \(9\le3\sum a^2\), so \(\sum a^2\ge3\).
Checks equality at \(a=b=c=1\).