Let $C= \{x \in \mathbb{R^2}:x_1 \geq x_2^2\}$.
How to show that $C$ is a convex set?
I tried to use the definition and I got:
Let $x_1,u \in C$ and $t \in [0,1]$. Then:
($tx_1+(1-t)u)\geq x_2^2=tx_1+u-tu \geq x_2^2 \Leftrightarrow x_2 \leq \sqrt{(tx_1+u-tu)}$
Here I don't know how to continue. I'm not sure how to conclude that this convex combination is an element of $C$.
Let $x, y \in C$ and $t \in [0, 1]$.
By Cauchy-Schwarz inequality, we have $$\left[ (1 -t) x_{2} +t y_{2} \right]^{2} = \left[ \sqrt{1 -t} \left( \sqrt{1 -t} \, x_{2} \right) + \sqrt{t} \left( \sqrt{t} \, y_{2} \right) \right]^{2} \leq \left[ (1 -t) +t \right] \left[ (1 -t) x_{2}^{2} +t y_{2}^{2} \right] \, \text{,}$$ that is $$\left[ (1 -t) x_{2} +t y_{2} \right]^{2} \leq (1 -t) x_{2}^{2} +t y_{2}^{2} \, \text{.}$$
Now, since $x, y \in C$, we have $x_{2}^{2} \leq x_{1}$ and $y_{2}^{2} \leq y_{1}$, and hence $$\left[ (1 -t) x_{2} +t y_{2} \right]^{2} \leq (1 -t) x_{1} +t y_{1} \, \text{,}$$ that is $(1 -t) x +t y \in C$.