I'm looking for the proof of an inequality such that I can prove the problem below (this is not the problem I need an answer to, it is the inequality later on that I don't get).
Problem
$$\text{Let } V = \{ x \in \mathbb{R}^2: 1-x_1^2-x_2^2 \leq \alpha\} \text{ be a given set. Prove that for any } \alpha \in \mathbb{R} \text{, set V is convex.}$$
Solution
$$\text{For } \lambda \in [0,1] \text{ and } x,y \in V \text{, we want to show that } \lambda x+(1-\lambda)y \in V.$$
$$\text{Since } x,y \in V, \text{we know that } 1-x_1^2-x_2^2 \leq \alpha \text{ and } 1-y_1^2-y_2^2 \leq \alpha$$
$$\text{Now we want to show that } 1-[\lambda x_1 + (1-\lambda)y_1]^2-[\lambda x_2 + (1-\lambda)y_2]^2 \leq \alpha$$
$$ \text{By rewriting the equation above to }\lambda[1-x_1^2-x_2^2] + (1-\lambda)[1-y_1^2-y_2^2] \leq \lambda\alpha + (1-\lambda)\alpha = \alpha$$
$$\text{Since } \alpha, x \text{ and } y \text{ were given arbitrarily, this holds for any value of these variables.}$$
Actual question
In order to do this, the answer sheet states that the following inequality holds (similar for $x_2$ and $y_2$) $$\lambda^2 x_1^2 + 2\lambda(1-\lambda)x_1y_1 + (1-\lambda)^2y_1^2 \leq \lambda x_1^2 + (1-\lambda)y_1^2$$ How can I derive this inequality? How can I show that this inequality holds?
EDIT: I know that $2xy \leq x^2+y^2$, inserting this gives $\lambda^2 x_1^2 +\lambda(1-\lambda)(x_1^2+y_1^2)+(1-\lambda)^2y_1^2$. I don't see how this is less than $\lambda x_1^2 + (1-\lambda)y_1^2$ since $\lambda(1-\lambda)(x_1^2+y_1^2) \geq 0$?