<s>[INST] Variation on Pythagoras: If $a^2 + b^2 = c^2$, then $a + b \leq c\sqrt{2}$ This is a very interesting word problem that I came across in an old textbook of mine. So I know its got something to do with derivative of the Pythagorean Theorem using calculus, trigonometry, geometry, or plain old algebra, which yields the shortest, simplest proofs, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes:

Prove that for $a, b, c > 0$,
if $a^2 + b^2 = c^2$, then $a + b \leq c\sqrt{2}$

 [/INST] 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: 

$a,b,c$ 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.
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