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.
 A: Edit: perhaps this is what you are thinking of?
Fix $S>0$ and suppose that $2b<S$. With $b>0$, this is equivalent to $4b^2<S^2$ or $b^2\in(0,\frac{S^2}{4})$. Then a counter example is guaranteed if we can show that for any $P\in(0,\frac{S^2}{4})$, there exist $a>0,c>0$ such that $a+c=S$ and $ac=P$. Such $a$ and $c$ are roots to the quadratic polynomial
$$
x^2-Sx+P.
$$
With $0<P<S^2/4$, you can use the quadratic formulas to check that this polynomial indeed has 2 positive roots. Because you can just pick $P>b^2$ (but still less than $S^2/4$), this proves that $a+c>2b$ does not guarantee that $ac>b^2$. But like others have said, a counter example is the best method.

Original answer:
Another counter example: $7+1>2\times 3$ but $7\times 1<3^2$. This one and yours share a common trait that $|a-c|$ is relatively large compared to the smaller of the two.
In general, given $a+c$ and $a,c>0$, to make "small" $ac$, try increasing $|a-c|$. This follows from $4ac = (a+c)^2-(a-c)^2$.
A: 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.
