Problem
ALG-B2-M07-P016 Return the scale
#16
★★★★★ Level 5 of 5
Suppose that for all \(x,y,z\ge0\) with \(x+y+z=1\), it is proved that \(x^2+y^2+z^2\ge\frac13\). Deduce for all \(a,b,c\ge0\): \[a^2+b^2+c^2\ge\frac{(a+b+c)^2}{3}.\]
Hint. Take \(x=\frac{a}{a+b+c}\), \(y=\frac{b}{a+b+c}\), \(z=\frac{c}{a+b+c}\).
If \(a+b+c=0\), the result is obvious. Otherwise set \(S=a+b+c\), \(x=a/S\), \(y=b/S\), \(z=c/S\). Then \(x+y+z=1\), so \(\frac{a^2+b^2+c^2}{S^2}\ge1/3\). Multiply by \(S^2\).
A very important formal skill: returning from normalization.