how to show that $a+c>2b$ do not guarantee that $ac>b^2$? HW problem. I solved it by giving counter example. But it will be nice if there is an algebraic way to show this.  
Given both $a,c$ are positive, and given that $a+c>2b$, the question asks if this guarantee that $ac>b^2$. The answer is no. Counter example is $a=8,c=2,b=4$. I tried to show this using algebra, but my attempts are not leading to anything. I started by squaring both sides of $a+c>2b$ but this lead to nothing I can see.
I thought someone here would know of a smart easy way to do this.

Given that $a\gt 0$ and $c\gt 0$, then
$$ a +c\gt 2b \Rightarrow \frac{a+c}{2} \gt b $$
And
$$ ac \gt b^2 \Rightarrow \sqrt{ac} \gt b $$
Then by AM-GM, we have 
$$ \frac{a+c}{2} \ge \sqrt{ac} \gt b $$
However, if we let $a=16$, $c=1$ and $b=5$
$$ \frac{16+1}{2} \ge \sqrt{16\cdot 1} \gt 5 $$
$$ 8.5 \ge 4\gt 5 $$
Which is clearly not true and thus a contradiction.
