> 
> Find the maximum value of $\sqrt{x - 144} + \sqrt{722 - x} .$
> 
> 
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**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?
['https://math.stackexchange.com/questions/3814894', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/772237/']
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 $a=b$.
Use

$$2(p^2+q^2)-(p+q)^2=\cdots\ge0$$

$$\implies p+q\le\sqrt{2(p^2+q^2)}$$

Here $p^2=x-144,q^2=722-x$