If x+y+z=c , show that $\displaystyle x^2+y^2+z^2\geq\frac{1}{3}c^2$
If x+y+z=c , show that $\displaystyle x^2+y^2+z^2\geq\frac{1}{3}c^2$
This follows from the Cauchy–Schwarz inequality. Applied to the vectors (x,y,z) and (1,1,1), the C–S inequality says that $\displaystyle (x+y+z)^2\leqslant (x^2+y^2+z^2)(1^2+1^2+1^2)$, or in other words $\displaystyle c^2\leqslant3(x^2+y^2+z^2)$.