Is $(tr(A))^n\geq n^n \det(A)$ for a symmetric positive definite matrix $A\in M_{n\times n} (\mathbb{R})$ If $A\in M_{n\times n} (\mathbb{R})$ a positive definite symmetric matrix, Question is to check if :
$$(tr(A))^n\geq n^n \det(A)$$
What i have tried is :
As $A\in M_{n\times n} (\mathbb{R})$ a positive definite symmetric matrix, all its eigen values would be positive.
let  $a_i>0$ be eigen values of $A$ then i would have :
$tr(A)=a_1+a_2+\dots +a_n$ and $\det(A)=a_1a_2\dots a_n$
for given inequality to be true, I should have $(tr(A))^n\geq n^n \det(A)$ i.e., 
$\big( \frac{tr(A)}{n}\big)^n \geq \det(A)$
i.e., $\big( \frac{a_1+a_2+\dots+a_n}{n}\big)^n \geq a_1a_2\dots a_n$
I guess this should be true as a more general form of A.M-G.M inequality saying
$(\frac{a+b}{2})^{\frac{1}{2}}\geq ab$ where $a,b >0$ 
So, I believe $(tr(A))^n\geq n^n \det(A)$ should be true.. 
please let me know if i am correct or try providing some hints if i am wrong.
EDIT : As every one say that i am correct now, i would like to "prove" the result which i have used just like that namely generalization of A.M-G.M inequality..
I tried but could not see this result in detail. SO, i would be thankful if some one can help me in this case.

This is really a Calculus problem! Indeed, let us look for the maximum of $h(x_1,\dots,x_n)=x_1^2\cdots x_n^2$ on the sphere $x_1^2+\cdots+x_n^2=1$ (a compact set, hence the maximum exists). First note that if some $x_i=0$, then $h(x)=0$ which is obviously the minimum. Hence we look for a conditioned critical point with no $x_i=0$.
For this we compute the gradient of $h$, namely
$$
\nabla h=(\dots,2x_iu_i,\dots),\quad u_i=\prod_{j\ne i}x_j^2,
$$
and to be a conditioned critical point (Lagrange) it must be orthogonal to the sphere, that is, parallel to $x$. This implies $u_1=\cdots=u_n$, and since no $x_i=0$ we conclude $x_1=\pm x_i$ for all $i$. Since $x$ is in the sphere, $x_1^2+\cdots+x_1^2=1$ and $x_1^2=1/n$. At this point we get the maximum of $h$ on the sphere:
$$
h(x)=x_1^{2n}=1/n^n.
$$
And now we can deduce the bound. Let $a_1,\dots,a_n$ be positive  real numbers and write $a_i=\alpha_i^2$. The point $z=(\alpha_1,\dots,\alpha_n)/\sqrt{\alpha_1^2+\cdots+\alpha_n^2}$ is in the sphere, hence
$$
\frac{1}{n^n}\ge h(z)=\frac{\alpha_1^2\cdots\alpha_n^2}{(\alpha_1^2+\cdots+\alpha_n^2)^n}=\frac{a_1\cdots a_n}{(a_1+\cdots+a_n)^n},
$$
and we are done. 
