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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.

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    $\begingroup$ Good observation. And yes, you are correct. $\endgroup$ – Shuchang Nov 5 '13 at 11:33
  • $\begingroup$ I see no reason why your argument could be wrong. $\endgroup$ – Han de Bruijn Nov 5 '13 at 11:39
  • $\begingroup$ 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.. please help me to see that... $\endgroup$ – user87543 Nov 5 '13 at 11:49
  • $\begingroup$ There are some hints and links at math.stackexchange.com/questions/483842/… $\endgroup$ – Gerry Myerson Nov 5 '13 at 11:52
  • $\begingroup$ @GerryMyerson : Thank you Sir. $\endgroup$ – user87543 Nov 5 '13 at 11:54
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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.

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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}

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