We solve this derivative of the Pythagorean Theorem using calculus, trigonometry,
geometry, or plain old algebra, which yields the shortest, simplest proofs!
Calculus:
The heavy hand of calculus does not provide a short
or elegant proof, but I always like seeing more than one approach.
$a^2 + b^2 = c^2$
$a^2 = c^2 - b^2 $
$a = \sqrt{c^2 - b^2} $
$a + b = \sqrt{c^2 - b^2} + b $
To maximize $a + b$, we take its derivative and set it to $0$.
Derivative is $$\frac{(\sqrt{c^2 - b^2} - b)}{\sqrt{c^2 - b^2}} $$
Setting that to $0$, we have
$\sqrt{c^2-b^2} = b$
$c^2 - b^2 = b^2 $
$c^2 = 2b^2$
We can also reach this point by seeing
that the derivative is also equal to $\frac{a - b}{a}
Setting $a - b = 0$, we have $a = b$,
$b = a = \frac{c}{\sqrt2} = \sqrt2 \times \frac c2$
At the maximum, $a + b = \sqrt2 \times c$,
therefore $a + b \leq \sqrt2 c$.
Trigonometry:
$a = c$ $\cos t$ and $b = c \sin t$
So $a + b = c \times ( \cos t + \sin t)$
As $c$ is constant, we need to maximise $\cos t + \sin{t}$
This is of the form $a \cos t + b \sin t$
$a = 1$ and $b = 1$
let $a = 1 = r \cos a$
B = 1 = r sin a
square and add: r = sqrt(2) and tan a = 1
so 1 cos t + 1 sin t
$= r \cos a \cos t + r \sin a \sin t $
$= r \cos(t-a)$
$= \sqrt2 \cos \times\ (t- \frac{\pi}{4})$
$a+b = \sqrt2 \cos (t-\frac{\pi}{4})\times c$
so $\leq \sqrt2 \times C$ as $\cos(t- \frac{\pi}{4}) \leq 1$
And the simplest proofs follow.
Geometry:
![enter image description here](https://cdn.statically.io/img/i.sstatic.net/8znxs.png)
$a,b,c$ are the sides of a right triangle.
Since $a, b$, and $c$ are positive real numbers:
$a+b > \sqrt2 \times c \geq (a+b)^2 > 2c^2 \geq 2ab > c^2$.
(since $a^2 + b^2 = c^2$)
The triangle $ABC$ is a right triangle, so it can be
inscribed in a circle with diameter $c$.
The height of this triangle, $h, \leq \frac c2$ (radius of circle).
The area of triangle = $\frac{ch}{2} = \frac{ab}{2}$.
$h \leq \frac c2$ (radius of circle).
$h² \leq \frac{ch}{2} \leq (\frac c2)^2 = \frac{c^2}{4} $
So $\frac{ab}{2} \leq \frac{c^2}{4}$
and $2ab \leq c^2$
Algebra:
We obviously have that: $0 \leq (a - b)^2$
$0 \leq a^2 - 2ab + b^2$
$a^2 + 2ab + b^2 \leq 2a^2 + 2b^2 = 2c^2 $
$(a + b)^2 \leq 2c^2$
EDIT: I should have just written:
$(a + b)^2 \leq (a - b)^2 + (a + b)^2 = 2c^2$
It would have been a nice one liner.