Finding Max/Min of a function in three variables, subject to two constraints 
Find the maximizers or minimizers of
  $$f(x_1,x_2,x_3) = \frac{1}{x^2_1 + x^2_2 + x^2_3}$$
  with the given constraints
  $$h_1(x_1,x_2,x_3) = 1 - x^2_1 - 2x^2_2 - 3x^2_3 = 0$$
  and
  $$h_2(x_1,x_2,x_3) = x_1 + 2x_2 + x_3 = 0.$$

My solution:
I tried to solve using Lagrange multipliers: thus coming up with these 3 equations, that we want to solve for $\lambda_1$ and $\lambda_2$
$$\begin{cases}
\dfrac{-2x_1}{(x^2_1 + x^2_2 + x^2_3)^2} = -2x_1 \lambda_1 + \lambda_2 \\
\dfrac{-2x_2}{(x^2_1 + x^2_2 + x^2_3)^2} = -4x_2 \lambda_1 + 2\lambda_2 \\
\dfrac{-2x_3}{(x^2_1 + x^2_2 + x^2_3)^2} = -6x_3 \lambda_1 + \lambda_2 
\end{cases}$$
But trying to solve this - is just insane. I cannot produce any results at all. There must be a trick here or something? Perhaps another way of solving.
 A: Hint. Finding the maximizers (minimizers) of
$$f(x_1,x_2,x_3) = \frac{1}{x^2_1 + x^2_2 + x^2_3}$$
is the same of finding the minimizers (maximizers)  of
$$g(x_1,x_2,x_3)=x^2_1 + x^2_2 + x^2_3.$$
P.S. Your problem has a nice geometric interpretation.
The first constraint is an ellipsoid centered at the origin, whereas the second constraint is a plane through the origin. Hence the domain of $f$ is the ellipse given by the intersection of the ellipsoid and the plane. Such ellipse is centered at the origin. Moreover $f$ is the the inverse of the square of the euclidean distance of a point of the ellipse from the origin. So the minimal (maximal) value of $f$ should be the inverse of the square of semi-major axis (semi-minor axis). 
A: Hint.
You found
$$
\begin{cases}
\dfrac{-2x_1}{(x^2_1 + x^2_2 + x^2_3)^2} = -2x_1 \lambda_1 + \lambda_2 \\
\dfrac{-2x_2}{(x^2_1 + x^2_2 + x^2_3)^2} = -4x_2 \lambda_1 + 2\lambda_2 \\
\dfrac{-2x_3}{(x^2_1 + x^2_2 + x^2_3)^2} = -6x_3 \lambda_1 + \lambda_2 
\end{cases}
$$
Calling
$$
\mu_1 = \lambda_1(x^2_1 + x^2_2 + x^2_3)^2\\
\mu_2 = \lambda_2(x^2_1 + x^2_2 + x^2_3)^2
$$
you can follow with
$$
\begin{cases}
-2x_1= -2x_1 \mu_1 + \mu_2 \\
-2x_2 = -4x_2 \mu_1 + 2\mu_2 \\
-2x_3 = -6x_3 \mu_1 + \mu_2 
\end{cases}
$$
now solving for $x_1,x_2,x_3,\mu_1$
$$
\begin{cases}
-2x_1= -2x_1 \mu_1 + \mu_2 \\
-2x_2 = -4x_2 \mu_1 + 2\mu_2 \\
-2x_3 = -6x_3 \mu_1 + \mu_2 \\
x_1+2x_2+x_3=0
\end{cases}
$$
we obtain
$$
\left[
\begin{array}{cccc}
x_1 & x_2 & x_3 & \mu_1\\
\frac{1}{2} \left(4+\sqrt{6}\right) \mu _2 & \left(1-\sqrt{6}\right) \mu _2 & \frac{1}{2} \left(-8+3
   \sqrt{6}\right) \mu _2 & -\frac{1}{10} \left(6+\sqrt{6}\right) \\
 \frac{1}{2} \left(4-\sqrt{6}\right) \mu _2 & \left(1+\sqrt{6}\right) \mu _2 & -\frac{1}{2} \left(8+3
   \sqrt{6}\right) \mu _2 & \frac{1}{10} \left(-6+\sqrt{6}\right) \\\end{array}
\right]
$$
and substituting into $1 - x^2_1 - 2x^2_2 - 3x^2_3 = 0$ we obtain finally
$$
\cases{-250 \left(54+19 \sqrt{6}\right) \mu _2^2+342 \sqrt{6}+847=0\\
250 \left(19 \sqrt{6}-54\right) \mu _2^2-342 \sqrt{6}+847=0
}
$$
and thus, the solutions for the stationary points $x_1^*,x_2^*,x_3^*$
