# Algebra question help here please? [on hold]

Let there be $u=(a,b)$ and $v= (1;1)$. Using Schwarz inequality prove that $[(a+b)/2]^2 = (a^2+b^2)/2$.

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## put on hold as off-topic by 6005, Normal Human, N. F. Taussig, Michael Albanese, LeucippusNov 22 at 0:29

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And the identity you ask us to prove is false. – Did Dec 7 '12 at 6:21
This is on my textbook. I know I have to use || u+v|| =sqrt((u+v)*(u+v)) but what do I do now? – Nottobe Dec 7 '12 at 6:24
Did you want to write inequality instead of equality? The result in you post is false - try $a=0$ and $b=1$. – Martin Sleziak Dec 7 '12 at 6:25
No ,it is exactly like this in my textbook, – Nottobe Dec 7 '12 at 6:26

Hint: It's straight plug in. Left side of usual Cauchy-Schwarz Inequality: $(a\cdot 1+b\cdot 1)^2$. Right side: $(a^2+b^2)(1^2+1^2)$. Manipulate a little.
First of all, it is not equal. When you substitute in the Schwartz Inequality you get $(a+b)^2\le (a^2+b^2)(2)$. Divide both sides by $4$, as you suggested, and we get $\left(\frac{a+b}{2}\right)^2 \le \frac{a^2+b^2}{2}$. If you were told to prove equality, it was a typo, take $a=1$, $b=3$, we don't get equality. Please do not write $=$, it is wrong. – André Nicolas Dec 7 '12 at 6:37