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How do I prove this inequality? In the task before, I had to prove the Cauchy-Schwarz inequality by using complex numbers but I'm kind of lost here.

for $a,b \ge 0$, $\;(a+b)/2 \le=\sqrt{(a^2+b^2)/2}$

Any help would be very appreciated!

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closed as off-topic by Did, José Carlos Santos, Ethan Bolker, Moishe Kohan, ಠ_ಠ Nov 13 '17 at 23:41

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$(a-b)^2 \geq 0 \rightarrow a^2-2ab+b^2 \geq 0 \rightarrow 2a^2+2b^2-(a+b)^2 \geq 0$ which means $$\frac{(a+b)^2}{4} \leq \frac{a^2+b^2}{2}$$ and finally take the square root.

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Since you proved Cauchy-Schwarz before, apply it to the vectors $\;\bigl(\frac12,\frac12\bigr)$ and $(a,b)$.

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