$\sqrt{x}+\sqrt{2-x}\geq a$ has a solution with interval $\leq 3$. [Solve for "a"].

Question

I need to solve this inequality. The first part of the problem was to solve for $a=3$ implies $x\in [0,2]$. This makes me think that I will somehow use this. I try to find the solution for the second part of the problem by squaring the inequality. I get $2\sqrt{(-x^2+2x)}\geq a-2$. This is equal to two systems: $$\begin{cases} a<2\\ x\in [0,2] \end{cases}\quad\mbox{and}\quad \begin{cases} a\geq2\\ 4x^2-8x+a^2-4a+2\geq 0 \end{cases} $$

What do I do know and is this actually correct? Please excuse me for this awful editing but I am on mobile. Also, there is no tag parametrical radical inequalities.

EDIT: I continued trying with the second system by finding the roots distance with the formula $\sqrt{D}/|a|$ but with no results.

Answer 1

Note that the function $f(x):=\sqrt{x}+\sqrt{2-x}$ is defined in $[0,2]$, it is symmetric with respect to $x=1$, i. e. $f(1+x)=f(1-x)$, and it is strictly increasing in $[0,1]$ (because $f'>0$ in $(0,1)$). Hence $f$ attains its maximum value $2$ at $1$ and its minimum value $\sqrt{2}$ at the endpoints $0$ and $2$.

It follows that the inequality $$\sqrt{x}+\sqrt{2-x}\geq a$$ has at least a solution if and only if $a\leq 2$ and the set of solutions is the interval $$\left[1-\frac{a}{2}\sqrt{4-a^2},1+\frac{a}{2}\sqrt{4-a^2}\right]$$ when $a\geq \sqrt{2}$, and it is the whole domain $[0,2]$ if $a\leq \sqrt{2}$.

Answer 2

Hint:

As $a$ is real, $0\le x\le2$

WLOG $x:2\sin^2t,0\le t\le\dfrac\pi2$

Now $\sin t+\cos t=\sqrt2\cos(45^\circ-t)$

Answer 3

If $y=\sqrt x+\sqrt{2-x},$

$y^2=x+2-x+2\sqrt{x(2-x)}$

Now $\dfrac{x+2-x}2\ge\sqrt{x(2-x)}$ as $\sqrt p\ge0$ for real $p$