<s>[INST] If $-1 \le x \le 1$, what is the maximum value of $x+\sqrt{1-x^2}$? (Cannot use calculus method) If $-1 \leq x \leq 1$, what is the maximum value of $x+\sqrt{1-x^2}$? (Cannot use calculus method)
As stated in the problem, I can't use calculus. Therefore, I'm using things I've learnt so far instead:
One of the things I have tried most successfully is using trig substitutions...
For example, if I substitute $x = \sin \phi$, this yields $\sin \phi + \cos \phi$
But what should I do next? Or there are any methods else to solve the problem?
 [/INST] We have
$$(x + \sqrt{1 - x^2})^2
+ (x - \sqrt{1 - x^2})^2
= 2x^2 + 2(1 - x^2) = 2 \tag{1}$$
which results in
$$x + \sqrt{1 - x^2} \le \sqrt 2.$$
(Note: Alternatively, we my use $(a + b)^2 \le 2(a^2 + b^2)$ to get
$(x + \sqrt{1 - x^2})^2 \le 2[x^2 + (1 - x^2)] = 2$.)
Also, when $x = 1/\sqrt 2$, we have $x + \sqrt{1 - x^2} = \sqrt 2$.
(Note: From (1), letting $(x - \sqrt{1 - x^2})^2 = 0$, we get $x = \sqrt{1 - x^2}$ or $x = 1/\sqrt 2$.)
Thus, the maximum of $x + \sqrt{1 - x^2}$ on $[-1,1]$ is $\sqrt 2$.
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