Is the smallest Pythagorean triple angle $\sim 4.9^\circ$?

Question

Pythagorean theorem $$a^2+b^2=c^2$$

Where $(a,b,c)$ are Pythagorean triples.

We are claiming that this inequality hold for all Pythagorean triples, $${3-2\sqrt{2}\over 2}\le{ab\over c^2}$$

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Let examine this inequality

$$3-2\sqrt{2}\le2\sin{\theta}\cos{\theta}$$

$$4.9396...^\circ\le \theta$$

So what this is saying is that the smallest angle of a Pythagoras triples can't go below $4.9396^\circ$

How can we show that this is true or false?

Answer 1

Let $a= 999999999999$, $b=2000000$ and $c=1000000000001$. This is a Pythagorean triangle. But $\arcsin(b/c) = 2\times 10^{-6}.$

Answer 2

Not a chance for a nonzero minimum angle. Suppose we render

$(2n+1)^2=(2n^2+2n+1)+(2n^2+2n)$.

Since the terms on the right differ by one unit the difference of squares factorization gives

$(2n+1)^2=(2n^2+2n+1)^2-(2n^2+2n)^2$.

And then

$(2n^2+2n+1)^2=(2n+1)^2+(2n^2+2n)^2$.

As $n$ is allowed to increase without bound the ratio of the smaller leg to the hypotenuse tends to a limit of zero and the angle opposite that leg does the same. Also, of course, the larger acute angle has no limit below $90°$.