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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.

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2 Answers 2

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

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Because by C-S: $$\frac{1}{2}a^2+\frac{1}{3}b^2+\frac{1}{6}c^2=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)\left(\frac{1}{2}a^2+\frac{1}{3}b^2+\frac{1}{6}{c^2}\right)\geq\left(\frac{1}{2}a+\frac{1}{3}b+\frac{1}{6}c\right)^2$$

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