For a Fixed Variance, Gaussian Distribution Maximizes Entropy? I was reading this paper. In page 5, second column, they mention that, 
$$h(Q) + h(P) \ge log(e \pi) => \sigma(Q) * \sigma(P) >= \frac{1}{2}$$
Where entropy $h$ is defined in the following way:
$$h(Q) = - \int_{-\infty}^{\infty} \Gamma(q) * log( \Gamma(q)) dq$$. 
Then it was mentioned that if $\Gamma(q)$ follows a Gaussian distribution, then for a fixed variance, the entropy $h(Q)$ will be maximized. They said it can be shown by variational calculus with Lagrange multipliers. Could anyone please give a hint how do I proceed to prove it?
 A: First, some properties of $\Gamma(q)$.  It is a probability distribution so
$$
\int_{-\infty}^\infty \Gamma(q) \,dq = 1
$$
We can without loss of generality assume we have shifted the variable $q$ such that the mean is zero, so that
$$
\int_{-\infty}^\infty q\, \Gamma(q)\, dq = 0
$$
We can rescale $q$ in a linear way so that the "fixed variance" is one.
$$
\int_{-\infty}^\infty q^2\, \Gamma(q)\, dq = 1
$$
Subject to those three constraints we wish to find an extremum of 
$$\int_{-\infty}^\infty \Gamma(q)\, \log(\Gamma(q))\,dq $$
We re-write the fixed variance constraint in such a way that we can ignore the first constraint until the end of the problem, and normalize the resulting function at the end.  Namely,
$$
\int_{-\infty}^\infty q^2\, \Gamma(q)\, dq = 
\int_{-\infty}^\infty \Gamma(q) \,dq 
$$
So we have Lagrange multipliers $\lambda_1, \lambda_2$ associated with the mean and variance, and out Euler equation is
$$
\frac\partial{d\Gamma} \left( \Gamma(q)\, \log(\Gamma(q)) \right)+\lambda_1 \frac\partial{d\Gamma}\left( q\, \Gamma(q)\right)+\lambda_2 \frac\partial{d\Gamma}\left( q^2\, \Gamma(q) -\Gamma(q)\right) =0
$$
or
$$
1+\log(\Gamma(q)) + \lambda_1q + \lambda_2 (q^2-1) = 0
$$
which says that any extremum must be of the form 
$$ \Gamma(q) = e^{\lambda_2 q^2+ (\lambda_1-\lambda_2)q + (1-\lambda_2)}
$$
and now we have to choose $\lambda_1$ and $\lambda_2$ such that the mean and variance will come out right (which is something like $\lambda_1 = \lambda_2 = -\sigma^2/2)$ and finally normalize the function because we had punted the constraint that the probability integrates to $1$.
