Lagrange multiplier does not find global minimum I want to minimize the function
$$f(r_1,r_2,r_3)=\frac{1}{6}(r_1^2+r_2^2+r_3^2)-(\frac{1}{r_1^3}+\frac{1}{r_2^3}+\frac{1}{r_3^3})$$
subject to the conditions:
$$g(r_1,r_2,r_3)=\frac{1}{3}(r_1^2+r_2^2+r_3^2)-3(\frac{1}{r_1^3}+\frac{1}{r_2^3}+\frac{1}{r_3^3})=0$$
$$r_1,r_2,r_3\geq0$$
The first thing I notice is that the inequality constraints are not active. So I set
$$\nabla f(r_1,r_2,r_3)=\lambda\nabla g(r_1,r_2,r_3)$$
which gives me $$\lambda=\frac{r_i^5+9}{2r_i^5+27}$$ for each $i=1,2,3$. Since $\lambda$ is the same for each equation, I start to inspect the new function $h(r)=\frac{r^5+9}{2r^5+27}$, which is strictly increasing on $(0,\infty)$. So in order to minimize $f$, $r_i$'s must be the same. Then I plug this into the equality constraint and find $f=(\frac{3}{2})^\frac{2}{5}3^\frac{-3}{5}$
However, this does not give me the global minimum, which is when $r_1+r_2=r_3$ and $r_1=r_2$. But in that case, $\lambda$ is not a constant for each $i$. I can prove that $f$ attains a global minimum on the set $\{
g=0,r_1,r_2,r_3>0\}$. What did I do wrong here?
 A: Calling
$$
L(r_k,\lambda) = \frac{1}{6}(r_1^2+r_2^2+r_3^2)-\left(\frac{1}{r_1^3}+\frac{1}{r_2^3}+\frac{1}{r_3^3}\right)-\lambda\left(\frac{1}{3}(r_1^2+r_2^2+r_3^2)-3\left(\frac{1}{r_1^3}+\frac{1}{r_2^3}+\frac{1}{r_3^3}\right)\right)
$$
the stationary points for this lagrangian are the solutions for
$$
0=\nabla L = \cases{9-27\lambda-(2\lambda-1)r_k^5, \ \ \{k=1,2,3\}\\
\frac{1}{3}(r_1^2+r_2^2+r_3^2)-3\left(\frac{1}{r_1^3}+\frac{1}{r_2^3}+\frac{1}{r_3^3}\right)
}
$$
now substituting the found $r_k^*(\lambda)$ into the restriction we get  $\lambda^* = \frac 25$ and then $r_k = 3^{\frac 25}$ so apparently we have only a stationary point. This point seems to be a global minimum.
A: I write $x$ instead of $r.$  Your question is equivalent to minimizing
$${1\over 18}(x_1^2+x_2^2+x_3^2) \tag1$$
subject to
$$x_1^2+x_2^2+x_3^2=9\big(  \frac{1}{x_1^3}+\frac{1}{x_2^3}+\frac{1}{x_3^3}   \big) \tag2 $$
Applying the Lagrange multiplier method to (1) and (2), we conclude that the minimum occurs when $x_1=x_2=x_3.$
Solving (2) under this condition gives $x_1=x_2=x_3=\sqrt[5]9.$ Thus, the minimum value is $1\over 2\sqrt[5]3$.
