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?
Best Answer
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 $a=b$.