Find the maximum value of $\sqrt{x - 144} + \sqrt{722 - x}$ 
Find the maximum value of $\sqrt{x - 144} + \sqrt{722 - x} .$

What I Tried :- Let me tell you first . What I Tried is absolutely silly , but you may check for it .
I thought AM-GM would do the trick and got :-
$\sqrt{x - 144} + \sqrt{722 - x} \geq 2 \sqrt{\sqrt{(x - 144)(722 - x)}}$
$\rightarrow \sqrt{x - 144} + \sqrt{722 - x} \geq 2 \sqrt{\sqrt{(-x^2 + 866x + 103968)}}$
From here I realised that there's no useful information I can find for $x$ .
Back to Another Attempt :- let $P = \sqrt{x - 144} + \sqrt{722 - x}$ . Then :-
$P^2 = 578 + 2\sqrt{(x - 144)(722 - x)}$
As $P \geq 0$ , we have that $P^2 \geq 578 \rightarrow P \geq 17\sqrt2 .$
This Attempt seemed reasonable and I thought I already found the solution, but then I realised that this is the minimum value of $P$ , and now I am hopeless .
Wolfram Alpha gives the answer to be $34$ , which can also be checked by Trial and Error of integer values , but that does not seem to be a proof of this problem .
Can anyone help me with this?
 A: By Cauchy-Schwartz:
$a=x-144,b=722-x$
$\sqrt{a}+\sqrt{b}\le \sqrt{(a+b)(1+1)}=\sqrt{1156}$
(You can read https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality)
Alternbatively
${(\sqrt{a}+\sqrt{b})}^2=a+b+2\sqrt{ab}$
recall by AM-Gm
$2\sqrt{ab}\le a+b$
thus
${(\sqrt{a}+\sqrt{b})}^2\le 2(a+b)$
So finally,
$\sqrt{a} + \sqrt{b} \le \sqrt{2(a+b)}$
Equality holds for $a=b$.
A: We have that $P^2=578+2\sqrt{(x-144)(722-x)}$
To maximize $P^2$ is to maximize $(x-144)(722-x)$, which is a downward parabola. By symmetry, it is maximized right between the two roots, i.e. at $x=\frac{(722+144)}{2}=433$
This also maximizes $P$ since it is always positive, so the maximum is $P$ evaluated at $433$, which yields $34$
A: Use
$$2(p^2+q^2)-(p+q)^2=\cdots\ge0$$
$$\implies p+q\le\sqrt{2(p^2+q^2)}$$
Here $p^2=x-144,q^2=722-x$
A: I agree with Evariste's answer, but would like to offer an alternative approach.  My approach begins with Evariste's conclusion that it is desired to maximize
$(x - 144)(722 - x) = -x^2 + x(866) - (144 \times 722).$
This is equivalent to trying to maximize
$-x^2 + x(866)$, where (presumably) $x$ is required to be in the interval [144, 722].
Suppose you pose the equation $-x^2 + x(866) = k.$ 
Then the question is what is the largest possible (real) value for $k$ 
that will generate at least one real root for $x.$
The above equation is equivalent to the equation $x^2 - x(866) + k = 0.$
This equation will have at least one real root if and only if 
$[(866)^2 - 4k] \;\geq\; 0.$
This means that the largest permissible value of $k$ is 
$\frac{1}{4} \times (866)^2.$
It is immediate that choosing this value for $k$ will cause 
$[(866)^2 - 4k]$ to equal 0.
This means that with this choice of $k$, the root(s) of 
$x^2 - x(866) + k = 0$ will be 
$x = (866/2)$.
By the above analysis, the OP's original expression should therefore be maximized by 
$x = 433.$
A: By C-S $$\sqrt{x-144}+\sqrt{722-x}\leq\sqrt{(1+1)(x-144+722-x)}=34.$$
The equality occurs for $(1,1)||(x-144,722-x),$ which says that we got a maximal value.
