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