> 
> 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?
['https://math.stackexchange.com/questions/3814894', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/772237/']
Use

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

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

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