For vectors in an Inner product space, would this be a valid proof for the triangle inequality?

Question

Let $V$ be an inner product space. Show that $$||x + y|| ≤ ||x|| + ||y||$$ for all x, y ∈ V

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By Pythagoras’ Theorem $$||x + y||^2 = ||x||^2 + ||y||^2$$

$$||x + y||^2 = (||x|| + ||y||)^2-2||x||||y||$$ $$||x + y||^2 \leq (||x|| + ||y||)^2$$ $$||x + y|| \leq ||x|| + ||y||$$

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The proof in our course uses Cacuhy-Schwarz Inequality. But I want to know if this is a valid proofing method?

EDIT:ok I see the issue now. As x and y might not be orthogonal, we cannot use Pythagoras's here. Thank you all.

Answer 1

$(\|x\|+\|y\|)^2=\|x\|^2+\|y\|^2+2\|x\|\|y\|\ge \|x\|^2+\|y\|^2+2<x,y>=\|x+y\|^2.$ Hence we have the triangular inequality.

Answer 2

No, that is not correct. You can only apply Pythagoras' theorem if $x$ and $y$ are orthogonal.

Answer 3

You can't use Pythagoras because that doesn't apply to arbitrary vectors. The triangle inequality is equivalent, by squaring, to $$\Vert x+y\Vert^2\le\Vert x\Vert^2+\Vert y\Vert^2+2\Vert x\Vert\Vert y\Vert,$$and from $\Vert u\Vert^2=\langle u|u\rangle$ this reduces to $\langle x,\,y\rangle+\langle y,\,x\rangle\le2\Vert x\Vert \Vert y\Vert$. So you'll have to use Cauchy-Schwarz or something close to it to finish the proof for general $x,\,y\in V$.