# How to prove $x^2+y^2\geq y^2$

Intuitive it's easy to see that $x^2+y^2\geq y^2$ is true, but I want to prove this.

My attempt:

I know $a^2\geq 0$ for all $a\in \mathbb R$.

I let $a=x-y$ so

\begin{align} &(x-y)^2=x^2-2xy+y^2\geq 0 \\ &\iff x^2+y^2\geq 2xy \end{align} But I'm stuck here, what is the next step?

A second question. If I instead have a strict inequality $x^2+y^2> y^2$, this is only valid if $x\neq 0$, right?

• The latest edit suggests you're looking for a proof that starts by deducing $x^{2} + y^{2} \geq 2xy$ for all real $x$, $y$. Unfortunately, there's no obvious way to get from this inequality to the desired inequality: The natural next step is "erase what you have so far, and proceed as in the existing answers". – Andrew D. Hwang Aug 5 '17 at 19:22

Note that $x^2\geq 0$, where equality holds if and only if $x=0$. Therefore $$x^2+y^2 \geq 0 + y^2 = y^2$$ and equality holds if and only if $x=0$.
$x^2 \geqslant 0$ for all $x \in \mathbb{R}$, with equality for $x=0$.