Is it true that, for any Pythagorean triple $4ab > c^2$? 
Is it true that, for any Pythagorean triple $4ab > c^2$?

So this came up in a proof I was working on and it seems experimentally correct from what I've tried and I would imagine the proof is similar to proving,
$$ab < \frac{c^2}{2}$$
The idea I have for this approach is,
$$4ab > c^2$$
$$4ab > a^2 + b^2$$
(Then maybe something with the triangle inequality?)

So is this statement true (it seems to be), and how can I prove it?
A counter-example would also be acceptable.

 A: $15^2 + 112^2 = 113^2$, but $4 \cdot 15 \cdot 112 \le 113^2$.
A: If $a^2 + b^2 = c^2$ then 
$c^2 < 4ab$ is the same thing as claiming $a^2 + b^2 < 4ab$
This is true if and only if $a^2 - 2ab + b^2 < 2ab$ which is to say
$(a-b)^2 < 2ab$
There shouldn't be any reason that should be true in general.  If let $m = $ average/midpoint of $a,b$, that is, $m = \frac {a+b}e$ and $e = $ the radius of the interval $a$ to $b$, that is $e = |m-a| =|b-m| = \frac{|a-b|}2$ then we have 
$(a-b)^2 < 2ab \implies$
$4e^2 < 2(m-e)(m+e) = 2(m^2 - e^2)\implies$
$3e^2 < m^2$
Which utterly need not be true!
Of course finding a pythagorean triplet where $a= m\pm e$ and $b=m\mp e$ are integers so that $a^2 + b^2 = 2(m^2 + e^2)$ is not merely and integer but a perfect square may put a wrinkle in finding a counterexample.
But not an insurmountable wrinkle.
We need, for a counter-example, $3e^2 \ge m^2$ which just require $e$ be significantly large, which is utterly irrelevent (it would seem) to $2(m^2 + e^2)$ being a perfect square.
... But to find a counter example:
.....
Back to square 1:  We can find a Pythogorian triple $a^2 + b^2 = c^2$ by letting $b$ be any odd integer and if $b^2 = 2a+1$ so $c^2 = (a+1)^2$ ... retrosolving by letting $a = \frac {b^2-1}2$.
So we want a counter example of $c^2 = a^2 + b^2 = a^2 + 2a + 1 = (a+1)^2 \ge 4ab$ or substitutint $a = \frac {b^2-1}2$, we want a case where
$(\frac {b^2+1}2)^2 \ge 4\frac{b^2 -1}2b$.
Surely we can find a counter example.
$(b^2 + 1)^2 \ge 8(b^2-1)b$
$b^4 + 2b^2 + 1 \ge 8b^3 -8b$
$b^4 -8b^3 +2b^2 - 8b + 1 \ge 0$
yeah that can be solved...
(Obviously as $b\to \infty$ the $b^4-8b^3 + 2b^2 - 8b + 1\to \infty$ so there is some $K$ where $b > K$ will always have $b^4 -8b^3 +2b^2 - 8b + 1\ge 0$.) 
Taking a sledgehammer and letting $b > 8$, say, $b=9$ and $a = \frac {b^2 -1}2=40$ so $40^2 + 9^2 = 41^2> 4*9*40=36*40$
Yeah... it's not true.
(Actually our sledgehammer was fairly precise.  $b^4 -8b^3 +2b^2 - 8b + 1$ has only two real solutions; one between $0$ and $1$ (far closer to $0$ than to $1$) and the other between $7$ and $8$.  So of primmitive triplets where $b$ is odd and $b^2 = 2a+1$. $a = 40, b=9, c=41$ is the smallest primitive triplet (and thus smallest of all triplets) where $c^2 > 4ab$.)
( so $c^2 < 4ab \iff \frac {c}{\gcd(a,b,c)} < 41$)
