<s>[INST] Let $x,y \ge 0 $.Find the maximum value of $\cfrac{(3x+4y)^2}{x^2+y^2}$ Let $x,y \ge 0 $.Find the maximum value of $\cfrac{(3x+4y)^2}{x^2+y^2}$
This exercise is from the chapter of my book regarding the Cauchy-Schwarz inequality but I don't know if I've applied it correctly .
The problem asks for the maximum value of the expression so it's clear that it must be on the lesser side of the Cauchy Inequality ,i.e. 
\begin{array}
 RR&\ge \cfrac{(3x+4y)^2}{x^2+y^2} \\
R(x^2+y^2) &\ge (3x+4y)^2 \\
\left(\left(\sqrt{R}\right)^2 \right)  (x^2+y^2) &\ge (3x+4y)^2 \\
\end{array}
Now this would be true iff 
$$ \left(\left(\sqrt{R}\right)^2 +(4)^2\right)  (x^2+y^2) \ge (3x+4y)^2 $$ where $R=9$ ,which would imply that this is the maximum.
However I am not really sure if I am correct.
 [/INST] You are right till 
$$R (x^2+y^2) \ge (3x+4y)^2$$
Now comparing with the CS inequality
$$(3^2+4^2)(x^2+y^2) \ge (3x+4y)^2$$
$\implies R = 3^2+4^2 = 25$.
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