For an $n\times n$ positive definite matrix $A$, I wish to prove that 


  $$\det(A) \leq \bigg(\frac{Trace(A)}{n}\bigg)^n$$


To me this seems some form of AM-GM Inequality (Arithmatic Mean-Geometric Mean Inequality). Therefore If I can show the following, above inequality follows : 


  $$\det(A) \leq \prod_{i=1}^{i=n} A_{ii}$$


Any idea how to prove the above.
Thanks


Question: Weil's Riemann Hypothesis for dummies?
Answer: Here are the statements from Schmidt's book (as pointed to in my comment). 

(a) Suppose $f(x,y)$ is a polynomial of total degree $d$, with coefficients in the field of $q$ elements and with $N$ zeros with coordinates in that field. Suppose $f(x,y)$ is absolutely irreducible, that is, irreducible not only over the field of $q$ elements, but also over every algebraic extension thereof. Then $$|N-q|\le2g\sqrt q+c_1(d)$$ where $g$ is the genus of the curve $f(x,y)=0$. 

I am not up to explaining "genus" without algebraic geometry, but it is known that $g\le(d-1)(d-2)/2$, so if you are willing to settle for $$|N-q|\le(d-1)(d-2)\sqrt q+c_1(d)$$ then I think you have what you are after. 

(b) Let $\chi$ be a multiplicative character of order $d&gt;1$. Suppose that $f(x)$, a polynomial in one variable over the field of $q$ elements, has $m$ distinct zeros, and is not a $d$th power. Then $$\Bigl|\sum_{x\in{\bf F}_q}\chi(f(x))\Bigr|\le(m-1)\sqrt q$$