What is the maximum of $x + y + z$ when $x^2 + y^2 + z^2 = 8$?

What is the maximum of $x + y + z$ when $x^2 + y^2 + z^2 = 8$?

I came across this question today, but cannot get the right answer.

This is how I tried to solve it:
$$(x + y + z)^2 ≤ x^2 + y^2 + z^2 = 8 $$ $$x + y + z = \sqrt 8$$

Could someone please tell me where I went wrong? Thank you.

Solution 1:

Using Cauchy-Schwarz, we get:

$(x+y+z)^2 \leq (1+1+1)(x^2+y^2+z^2)$

and from here I believe you can work it out.

Also regarding your original inequality that you assumed, a counterexample happens when $(x, y, z) = (2, 2, 0)$, so might be that you either applied Cauchy incorrectly, or you had a wrong conceptual knowledge in the first place.

Solution 2:

Not an answer, but an illustration of the problem. The curves on the sphere are contours of $x+y+z$. Towards the upper right [in the direction of $(1,1,1)$], the contours have values $\{0,1,2,3,4,4.8\}$.