minimize $x^4 - 6x^2 y^2 + y^4$ given $x^2 + y^2 \leq 1$ I have a constrained optimization problem.  Can we maximize / minimize this function on the unit sphere?
$$  f(x,y,z) = x^4 - 6 x^2 y^2 + y^4 \quad\text{given that}\quad x^2 + y^2 + z^2 = 1$$ 
One idea could be to use the Cauchy-Schwartz inequality.  Since I forget the proof:
$$ (x^2 + y^2)^2 \geq 0  \text{ so that }x^4 + y^4 \geq 2 x^2 y^2 \text{ and }f(x,y,z) \geq - 4 x^2 y^2 \geq - 4 $$
I could try other rearrangements as well.  This one gives me an upper bound of $3$.
$$  x^4 - 6x^2 y^2 + y^4 \leq x^4 + 6x^2 y^2 + y^4 
= (x^2 + y^2)^2 + 4x^2 y^2 \leq 3\, \big( x^2 + y^2 \big)^2 \leq 
3\, \big( x^2 + y^2  + z^2 \big)^2  = 3$$
If I use some real analysis we know that the sphere as a subset of Euclidean space is compact, so that:
$$ -\infty < -4 \leq \min_{x^2 +y^2 + z^2 = 1} f(x,y,z) \leq  \min_{x^2 +y^2 + z^2 = 1} f(x,y,z) < 3 < +\infty$$
I'm trying to avoid Lagrange multipliers unless the're really natural here. Observer also that:
$$ \left[ x^2 + y^2  + z^2 =  1 \right] \to \left[ x^2 + y^2 \leq 1\right] $$
as the original problem was defined on the unit sphere but the $z$ is extraneous.

They might not be extraneous we could set spherical coordinates:
$$  (x,y,z ) = \big(\cos \theta \, \cos \varphi, \;\cos \theta \sin \varphi, \;\cos \varphi\big)$$
and we could put into our inequality:
\begin{eqnarray*} x^4 - 6x^2 y^2 + y^4 &=& \cos^4 \theta \cos^4 \varphi - 6 \cos^4 \theta \sin^2 \varphi \cos^2 \varphi+ \cos^4 \theta \sin^4 \varphi \\ \\
 &=& \cos^4 \theta \,\big( \cos^4 \varphi - 6 \cos^2 \varphi \sin^2 \varphi + \sin^4 \varphi \big)  \\ \\
&\leq & \cos^4 \varphi - 6 \cos^2 \varphi \sin^2 \varphi + \sin^4 \varphi
\end{eqnarray*}
This looks promising as I have reduced a three-dimensional problem to a problem with only an angle $\varphi$, but I may have lost something with the final "$\leq$" sign.
Just a tiny bit more:
$$
\cos^4 \varphi - 6 \cos^2 \varphi \sin^2 \varphi + \sin^4 \varphi
= (\cos^2 \varphi - \sin^2 \varphi)^2 - 4 \cos^2 \varphi \sin^2 \varphi
= \cos^2 2\varphi -  \sin^2 2\varphi
 $$
and if we use the double-angle identity.
$$ 1 \geq \cos^2 2\varphi -  \sin^2 2\varphi 
= \cos^2 2\varphi -  (1 -\cos^2 2\varphi)
= 2\, \cos^2 2\varphi - 1  \geq - 1$$
This is very similar to what I obtained before.
 A: Let $x=r \cos \theta, y = r \sin \theta$, then the cost reduces to
$r^4(\cos^4 \theta -6 \cos^2 \theta \sin ^2 \theta + \sin^4 \theta) = r^4 \cos(4 \theta)$.
Hence we can write an equivalent problem as extremising
$r^4 \cos(4 \theta)$ subject to $r \in [0,1]$.
A: Another approach would be to use cylindrical coordinates. Then your expression is $r^4\cdot g(\theta)$ with $0\leq r\leq1$, so understanding $g$'s extrema, together with extremal values of $0$ and $1$ for $r^4$ would tell you the extreme values of $f$. Specifically, 
$$\begin{align}
g(\theta)&=\cos^4(\theta)-6\cos^2(\theta)\sin^2(\theta)+\sin^4(\theta)\\
&=V^2-6V(1-V)+(1-V)^2\\
&=8V^2-8V+1\\
&=2(2V-1)^2-1
\end{align}$$
where $V=\cos^2(\theta)$ ranges from $0$ to $1$.
A: The first approach is good, except for its last step. The correct argument is
$$f(x,y,z)\ge -4x^2y^2\ge -(x^2+y^2)^2\ge -(x^2+y^2+z^2)^2 = -1.$$
Minimum attains at $x=y=\pm\sqrt2,z=0$.
A: A different approach. As you noted, being on or within the unit sphere just implies $x^2+y^2\leq 1$, and we can then ignore $z$.
$$F = x^4 - 6x^2 y^2 + y^4$$
$$F = (x^2-y^2)^2 - 4x^2 y^2$$
$$F = (x^2-y^2+2xy)(x^2-y^2+2xy)$$
Now substitute $x = r \cos\theta, y = r \sin\theta$. We know $0 \leq r\leq 1$.
$$F = (r^2\cos^2\theta-r^2\sin^2\theta+2r^2\cos\theta\sin\theta)
(r^2\cos^2\theta-r^2\sin^2\theta-2r^2\cos\theta\sin\theta)$$
$$F = r^4(\cos^2\theta-\sin^2\theta+2\cos\theta\sin\theta)
(\cos^2\theta-\sin^2\theta+2\cos\theta\sin\theta)$$
$$F = r^4[\cos(2\theta)+\sin(2\theta)][\cos(2\theta)-\sin(2\theta)]$$
$$F = r^4[\cos^2(2\theta) - \sin^2(2\theta)]$$
$$F = r^4\cos(4\theta)$$
Therefore
$$-1 \leq F \leq 1$$
with minimum $-1$ at
$$\theta = \frac{\pi}4, \frac{3\pi}4, \frac{5\pi}4, \frac{7\pi}4 \Rightarrow x= \pm \frac{1}{\sqrt{2}}, y = \pm \frac{1}{\sqrt{2}}$$
and maximum $1$ at
$$\theta = 0, \frac{\pi}2, \pi, \frac{3\pi}2 \Rightarrow x = \pm 1, y = 0 \mbox{ or } x = 0, y = \pm 1.$$
A: For $x=y=\frac{1}{\sqrt2}$ we get a value $-1$.
We'll prove that it's a minimal value.
We need to prove that
$$x^4-6x^2y^2+y^4\geq-1.$$ 
Indeed,
$$x^4-6x^2y^2+y^4+1\geq x^4-6x^2y^2+y^4+(x^2+y^2)^2=2(x^2-y^2)^2\geq0$$
and we are done!
