Proof of convexity from definition ($x^Tx$) I have to prove that function $f(x) = x^Tx, x \in R^n$ is convex from definition.
Definition: Function $f: R^n \rightarrow R$ is convex over set $X \subseteq dom(f)$ if $X$ is convex and the following holds: $x,y \in X, 0 \leq \alpha \leq 1 \rightarrow f(\alpha x+(1-\alpha) y)) \leq \alpha f(x) + (1-\alpha)f(y)$.
I got this so far:
$(\alpha x + (1-\alpha)y)^T(\alpha x + (1-\alpha)y) \leq \alpha x^Tx + (1-\alpha)y^Ty$
$\alpha^2 x^Tx + 2\alpha(1-\alpha)x^Ty + (1-\alpha)^2y^Ty \leq \alpha x^Tx + (1-\alpha)y^Ty$
I don´t know how to prove this inequality. It is clear to me, that $\alpha^2 x^Tx \leq \alpha x^Tx$ and $(1-\alpha)^2y^Ty \leq (1-\alpha)y^Ty$, since $0 \leq\alpha \leq 1$, but what about $2\alpha(1-\alpha)x^Ty$?
I have to prove this using the above definition.
Note: In Czech, the words "convex" and "concave" may have opposite meaning as in some other languages ($x^2$ is a convex function for me!). Thanks for any help.

$g(x) = \sqrt{x^Tx}$ is convex due to triangle inequality. And $h(x) = x^2$ is convex (one of the ways to see this is to use calculus).
$f(x) = h(g(x))$ and both of $h$ and $g$ are convex.
