# Where is this simple inequality in print?

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 $49<50$.

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Interpreting your numbers $a,b,c$ as sides of a right triangle, $\frac{a}{c}=\sin(\theta)$ and $\frac{b}{c}=\cos(\theta)$. Your inequality is equivalent to $$\sin(\theta)+\cos(\theta)=\sqrt{2}\cos\left(\theta+\frac{\pi}{4}\right)\le \sqrt{2}$$
Which is obviously true.

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It’s familiar in the slightly disguised (and generalized) form $\|x\|_1\le\sqrt n\|x\|_2$ in $\Bbb R^n$. I’d not expect it to have any surprising consequences: it’s really nothing more than the fact that squares of real numbers are non-negative.

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Given $\,(**)\;\;c^2=a^2+b^2\,$

$$\frac{a+b}{c}\leq\sqrt 2\Longleftrightarrow a^2+2ab+b^2\leq 2c^2\stackrel{\text{by}(**)}=2a^2+2b^2\Longleftrightarrow (a-b)^2\geq 0$$

and since the rightmost inequality is trivially true then the whole chain above is true.

This trivial proof seems to imply that either it exists in some book(s) or else it is so trivial that nobody cared about printing it unless used for some practical purpose

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This is an algebraic consequence of the fact that $x^2 \ge 0$ for $x \in \mathbb{R}$.

More specifically, let $a, b > 0$. Then: $$0 \le (a - b)^2$$ $$0 \le a^2 - 2ab + b^2$$ $$2ab \le a^2 + b^2$$ $$a^2 + 2ab + b^2 \le 2a^2 + 2b^2$$ $$(a + b)^2 \le 2a^2 + 2b^2$$ $$a + b \le \sqrt{2}\sqrt{a^2 + b^2}$$ $$\frac{a + b}{\sqrt{a^2 + b^2}} \le \sqrt{2}$$

For that reason, I do not believe that $(1)$ has any consequences that are not already implied by the simpler $x^2 \ge 0$ rule.

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So it's not even really about Pythagorean triples. EXCEPT that your final inequality doesn't usually have a rational number on the left side even when $a,b$ are integers. 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....... –  Michael Hardy Nov 18 '12 at 21:07
....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 $49<50$. –  Michael Hardy Nov 18 '12 at 21:08