If $a,b,c>0$ and $a^2+b^2=c^2$ then $ \frac{a+b}{c} \le\sqrt{2}.\tag{1} $ Equality holds iff $a=b$, so equality never holds for integral Pythagorean triples.
(If we write this as $(a+b)^2\le2c^2$, then for the Pythagorean triple $(3,4,5)$ this tells us that $\frac75<\sqrt{2}$, or in other words $49<50$, and for $(20,21,29)$ we deduce the result that $\frac{41}{29}<\sqrt{2}$, or $1681<1682$.)
Questions:
- Is $(1)$ "known" in the sense of being found in books/papers/etc.?
- Altough a proof of $(1)$ is trivial, might there be consequences that are not trivial?
Later edit: Maybe I should have asked which rational numbers $<\sqrt{2}$ are of this form.
I was drawing a crude illustration for purposes unrelated to the topic of this question. I plotted $(5,0)$, $(4,3)$, $(3,4)$, $(0,5)$ and their counterparts in other quadrants in order to draw a circle, and I wanted to draw the tangent line with slope $-1$ through $(\sqrt{2},0)$. I used $(7,0)$ to approximate that point and lo and behold, the line went straight through two of the points I'd plotted. So I realized that's because $7/5$ as an approximation to $\sqrt{2}$ errs on the small side. At least it gets me a way to prove that $49<50$.