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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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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. –  Did Dec 7 '12 at 6:20
    
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
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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
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1 Answer

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.

Check that you have typed correctly what you were asked to prove. If you were asked to prove the expressions are equal, well, they usually are not.

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so I have (a*1+b*1)^2 = (a^2+b^2)*2 I divide by 4,and I think I have proven it? –  Nottobe Dec 7 '12 at 6:30
    
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
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