Prove that $\frac{x^n+y^n}{2}$ Has a Local Minimum at $x=y=C^{\frac{1}{n}}$ with a Constraint $(\frac{x+y}{2})^n=C$ 
Question Prove that $\frac{x^n+y^n}{2}$ has a local minimum at $x=y=C^{\frac{1}{n}}$ with a constraint $(\frac{x+y}{2})^n=C$. (Try to use the method of Lagrange multipliers to solve this question.)


Using method of Lagrange multipliers, I got $x=y=C^{\frac{1}{n}}$, $\lambda=-1$.
And then I calculated the Hessian matrix of the Lagrange function at the point $x=y=C^{\frac{1}{n}}$ and $\lambda=-1$
$$\frac{n(n-1)}{2}C^{\frac{n-2}{n}}\begin{pmatrix} \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} \end{pmatrix}
$$
which is a positive-semidefinite matrix.
Here I don't know what to do next. In my course, I have learned the theorem

Theorem If the Hessian matrix of the Lagrange function at a point is positive-definite, the original function has a local minimum at this point.

I can draw a conclusion if it is a positive-definite matrix. But unluckily, it is a positive-semidefinite matrix. It seems to need further work to solve the question.
So my question is what should I do when I encounter a positive-semidefinite Hessian matrix. If possible, please tell me how does it work.

Geometrically the constraint $g(x,y)=C$ as a curve / submanifold of $\mathbb R^2$, the normal vector to the curve is $\nabla g$
$$
\nabla g = \frac n2(\frac{x+y}{2})^{n-1}[1,1]^T
$$
So the normal vector is always in the linear subspace $Span([1,1])$.
Then look at the Hessian matrix, $[1,1]$ is exactly the eigenvector with 0 eigenvalue, i.e. $Span([1,1])=null(H)$.
So when we restrict ourselves to the tangent space of the constraint manifold, $Span([1,1])^{\perp}=Span([1,-1])$, the Hessian matrix becomes positive definite. Then we can say we have a local minimum at $x=y=C^{\frac{1}{n}}$
