Problem 10.1. Consider the inequality $\sqrt{x}+\sqrt{2-x} \geq \sqrt{a}$, where $a$ is a real number.

a) Solve the inequality for $a=3$.

б) Find all $a$, for which the set of solutions of the inequality is a segment (possibly, a point) of length less than or equal to $\sqrt{3}$.

Kerope Chakaryan
10.1. a) For $a=3$ and $x \in[0,2]$ the inequality is equivalent to $2 \sqrt{x(2-x)} \geq$ 1, i.e. $4 x^{2}-8 x+1 \leq 0$. Hence its solutions are

$$
x \in\left[\frac{2-\sqrt{3}}{2}, \frac{2+\sqrt{3}}{2}\right]
$$

b) For $a \geq 0$ and $x \in[0,2]$ the inequality is equivalent to $2 \sqrt{x(2-x)} \geq$ $a-2$. If $a \leq 2$, then any $x \in[0,2]$ is a solution and the condition of the problem does not hold.

Let $a>2$. Then $4 x(2-x) \geq(a-2)^{2}$ (in particular, $\left.x \in[0,2]\right)$, i.e. $4 x^{2}-8 x+a^{2}-4 a+4 \leq 0$. It follows that $D=16 a(4-a) \geq 0$ and hence $a \in(2,4]$. In this case the solutions of the inequality are $x \in\left[x_{1}, x_{2}\right]$, where $x_{1} \leq x_{2}$ are the roots of the respective quadratic equation. The given condition becomes $x_{2}-x_{1} \leq \sqrt{3}$. Since $x_{2}-x_{1}=\frac{\sqrt{D}}{4}=\sqrt{a(4-a)}$, we obtain $a^{2}-4 a+3 \geq 0$. Taking into account that $a \in(2,4]$ we conclude that $a \in[3,4]$.