Part a)

Solution.

We have that for any \(x,y\in\mathbb{R},\) \[ 0 \leq (x-y)^2 = x^2 - 2xy + y^2,\] from which rearranging gives \[2xy \leq x^2 + y^2.\] Now, \[2xy = x^2 + y^2\; \Leftrightarrow\; 0 = (x-y)^2 \;\Leftrightarrow \;x-y = 0 \;\Leftrightarrow\; x=y.\] So equality holds if and only if \(x = y.\)