# The first part of the problem was to solve for a = 3 implies x &#x2208;<!-- ∈ -->

The first part of the problem was to solve for $a=3$ implies $x\in \left[0,2\right]$. 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 $2\sqrt{\left(-{x}^{2}+2x\right)}\ge a-2$.
This is equal to two systems:
$\left\{\begin{array}{l}a<2\\ x\in \left[0,2\right]\end{array}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left\{\begin{array}{l}a\ge 2\\ 4{x}^{2}-8x+{a}^{2}-4a+2\ge 0\end{array}$
What do I do know and is this actually correct?
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Christina Ward
Note that the function $f\left(x\right):=\sqrt{x}+\sqrt{2-x}$ is defined in $\left[0,2\right]$, it is symmetric with respect to $x=1$, i. e. $f\left(1+x\right)=f\left(1-x\right)$, and it is strictly increasing in $\left[0,1\right]$ (because ${f}^{\prime }>0$ in $\left(0,1\right)$). 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
$\sqrt{x}+\sqrt{2-x}\ge a$
has at least a solution if and only if $a\le 2$ 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 $a\ge \sqrt{2}$, and it is the whole domain $\left[0,2\right]$ if $a\le \sqrt{2}$.
###### Not exactly what you’re looking for?
Yahir Crane
Hint:
As $a$ is real, $0\le x\le 2$
WLOG $x:2{\mathrm{sin}}^{2}t,0\le t\le \frac{\pi }{2}$
Now $\mathrm{sin}t+\mathrm{cos}t=\sqrt{2}\mathrm{cos}\left({45}^{\circ }-t\right)$