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.

For convenience, we use the notation $A\succ 0$ to indicate that a symmetric matrix $A$ is positive definite. We can see the inequality $(tr(A))^n\geq n^n \det(A),\;\forall A\succ 0$ as
$$
\frac{1}{n}\mathrm{trace}(A)\geq \sqrt[n\,]{\det(A)}, \quad \forall A\succ 0.
$$
Note that, if $A=\mathrm{diag}(\lambda_1,\ldots,\lambda_i,\ldots,\lambda_n)$, that is, 
$$
A=
\begin{pmatrix}
\lambda_{1} & \cdots & 0           & \cdots & 0     \\
\vdots      & \ddots &\vdots       &        &\vdots \\
0           &\cdots  & \lambda_i   & \cdots & 0     \\ 
\vdots      &        &\vdots       &\ddots  &\vdots \\
0           &\cdots  &0            &\cdots  &\lambda_n
\end{pmatrix}
$$
we have 
$$
\frac{1}{n}\mathrm{trace}(A)\geq \sqrt[n\,]{\det(A)}
\Longleftrightarrow
\frac{\lambda_1+\ldots+\lambda_i+\ldots+\lambda_n}{n}\geq \sqrt[n\,]{\lambda_1\cdot\ldots\cdot\lambda_i\cdot\ldots\cdot\lambda_n}
$$
So we can see the inequality in question as a generalization of the inequality between the arithmetic mean and geometric mean. See a proof using forward–backward induction here.
For every $n\times n$ real symmetric matrix $A$, the eigenvalues are real and the eigenvectors can be chosen such that they are orthogonal to each other. Thus a real symmetric matrix $A$ can be decomposed as
$A=Q\Lambda {Q}^{T}  $
where $\Lambda$ is a diagonal matrix whose entries are the eigenvalues of $A$, and $Q$ is an orthonormal matrix. For a orthonormal matrix we have $Q^{-1}=A^{T}$ and $QQ^T=I$. With these observations the required inequality is the result of '$\det$', '$\mathrm{trace}$' properties and algebraic manipulations: 
\begin{align}
\frac{\mathrm{trace}(A)}{n}
=&
\frac{\mathrm{trace}(Q\Lambda Q^T)}{n}
\\
=&
\frac{\mathrm{trace}(Q^TQ\Lambda)}{n}
\\
=&
\frac{\mathrm{trace}(\Lambda)}{n}
\\
=&
\frac{\lambda_1+\ldots+\lambda_i+\ldots+\lambda_n}{n}
\\
\\
\geq&
\sqrt[n\,]{\lambda_1\cdot\ldots\cdot\lambda_i\cdot\ldots\cdot\lambda_n}
\\
=&
\sqrt[n]{\det(\Lambda)}
\\
=&
\sqrt[n]{\det(Q^TQ\Lambda)}
\\
=&
\sqrt[n]{\det(Q\Lambda Q^T)}
\\
=&
\sqrt[n]{\det(A)}
\end{align}
