Real Analysis – Application of the Cauchy-Schwarz Inequality

Question

Need to prove the following:
$(\frac{1}{2}a+\frac{1}{3}b+\frac{1}{6}c)^2\leq \frac{1}{2}a^2+\frac{1}{3}b^2+\frac{1}{6}c^2$
using the Cauchy-Schwarz inequality however direct application yields $(\frac{1}{2}a+\frac{1}{3}b+\frac{1}{6}c)^2\leq \frac{7}{18}(a^2+b^2+c^2)$ which is a dead end.

Answer 1

Hint: $\frac 12a+\frac 13b+\frac16c=\frac16a+\frac16a+\frac16a+\frac16b+\frac16b+\frac16c$