TITLE: Proof of convexity of $f(x)=x^2$
QUESTION [12 upvotes]: I know that a function is convex if the following inequality is true:
$$\lambda f(x_1) + (1-\lambda)f(x_2) \ge f(\lambda x_1 + (1-\lambda)x_2)$$ 
for $\lambda \in [0, 1]$ and $f(\cdot)$ is defined on positive real numbers.
If $f(x)=x^2$, I can write the following:
$$\lambda x_1^2 + (1-\lambda)x_2^2 \ge (\lambda x_1 + (1-\lambda)x_2)^2$$
$$0 \ge (\lambda ^2 - \lambda) (x_1^2 - x_2 ^ 2) $$
But I am not sure if this is true or not. How can I prove this?

REPLY [6 votes]: $$\lambda x_1^2 + (1-\lambda)x_2^2 \ge (\lambda x_1 + (1-\lambda)x_2)^2\iff$$
$$\lambda x_1^2+(1-\lambda)x_2^2\ge\lambda^2 x_1^2+2\lambda(1-\lambda)x_1x_2+(1-\lambda)^2x_2^2\iff$$
$$\lambda x_1^2(1-\lambda)+(1-\lambda)x_2^2(1-(1-\lambda))-2\lambda(1-\lambda)x_1x_2\ge 0\stackrel{\text{assuming}\;\lambda\neq1}\iff$$
$$\lambda x_1^2+\lambda x_2^2-2\lambda x_1x_2\ge 0\iff\lambda(x_1-x_2)^2\ge 0$$
and since the last inequality is obvious we're done (if $\;\lambda=1\;$ there's nothing to prove...)