The maximum and minimum of a function How can I find the maximum and minimum of $$L=\sqrt{1-x^2}+\sqrt{1-y^2}+\sqrt{1-z^2},$$
given that $x^2+y^2+z^2=1$?
I worked on some basic algebra, but it doesn't get me anywhere, so I'm stuck on it.
 A: By using lagrange multipliers $$ f(x,y,z) = \sqrt{1 - x^2} + \sqrt{1 - y^2} + \sqrt{1 -  z^2} $$ subject to $$ g(x,y,z) = x^2 + y^2 + z^2 - 1 = 0 $$
$$ F(x,y,z,\lambda) = f + \lambda g = \sqrt{1 - x^2} + \sqrt{1 - y^2} + \sqrt{1 -  z^2} + \lambda( x^2 + y^2 + z^2 - 1) $$
$$ F_x = \frac{-2x}{2\sqrt{1-x^2}} + 2\lambda x = 0 $$
$$ F_y = \frac{-2y}{2\sqrt{1-y^2}} + 2\lambda y = 0  $$
$$ F_z = \frac{-2z}{2\sqrt{1-z^2}} + 2\lambda z = 0  $$
$$ F_{\lambda} = x^2 + y^2 + z^2 - 1 = 0 $$
now solve this system of equations :-)
A: By Cauchy-Schwarz inequality
$$\sum_{cyc}\sqrt{1-x^2}\le\sqrt{3\cdot(1-x^2+1-y^2-z^2)}=\sqrt{6}$$
By the other hand,we have
\begin{align*}
&\sqrt{1-x^2}+\sqrt{1-y^2}+\sqrt{1-z^2}=\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{x^2+z^2}\\
&\ge\sqrt{0+y^2}+\sqrt{0+z^2}+\sqrt{y^2+z^2}\\
&=|y|+|z|+\sqrt{y^2+z^2}
\end{align*}
this problem since $y^2+z^2=1$,find the $|y|+|z|+1$ minum
let
$$|y|=a\ge 0,|z|=b\ge0, a^2+b^2=1$$
then
$$|y|+|z|+1=a+b+1=\cos{t}+\sin{t}+1=\sqrt{2}\sin{(t+\dfrac{\pi}{4})}+1\ge 2$$
A: Why don't you try to convert the equation in polar coordinates which will simplify the equation as $r=1$ meaning you only have two unknowns left to solve.
