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 $a + b \leq c\sqrt{2}$

 A: Proof without words:

(This space intentionally left blank.)
A: $(a+b)^2\\
\leq(a+b)^2+(a-b)^2\\
=a^2+2ab+b^2+a^2-2ab+b^2\\
=2a^2+2b^2
=2(a^2+b^2)
=2c^2$
Using the square root gives us:
$a+b\leq\sqrt{2}c$
A: You have $c=\sqrt{a^2+b^2}$, so your inequality is equivalent to
$$\frac{a+b}2 \le \sqrt{\frac{a^2+b^2}2}.$$
This is the well known inequality between quadratic and arithmetic mean. 
In this case we only need two variables, but it is true for more variables, too.

If $x_1,\dots,x_n\ge0$ are real numbers then
  $$\frac{x_1+x_2+\dots+x_n}n \le \sqrt{\frac{x_1^2+x_2^2+\dots+x_n^2}n}.$$
  The equality holds if and only if $x_1=x_2=\dots=x_n$.

Some links:


*

*Prove QM-AM inequality

*Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality at AoPS 

*Inequalities: A Mathematical Olympiad Approach By Radmila Bulajich Manfrino et al., p.36
A: Let us prove it by contradiction. So if the given condition is not true then $$ a+b> c\sqrt2 $$
$\because  a+b ,c\sqrt2>0$ so consider 
\begin{align}
 a+b   > c\sqrt2\\
&\implies& (a+b)^2 >  2c^2\\
&\implies& a^2+b^2+2ab>2c^2\\
&\implies& c^2+2ab-2c^2>0\\
&\implies& 2ab-c^2>0\\
&\implies& c^2-2ab<0\\
&\implies& a^2+b^2-2ab<0\\
&\implies& (a-b)^2<0
\end{align}
So we got a contradiction. Hence, $a+b\leq c\sqrt2. $
A: 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.
A: Assuming you are looking for $a+b\le c\sqrt 2$, as a hint, consider $(a+b)^2+(a-b)^2$
A: Let $x=(a,b)$ and $y=(1,1)$,  then we have:
$a+b=x\centerdot y\leq||x||\space||y||=\sqrt{a^2+b^2}\sqrt{2}=c\sqrt{2}$.
A: $(a-b)^2 + 2ab = c^2$. This gives $2ab \leq c^2$ since each summand is nonnegative.
$(a+b)^2 = c^2 + 2ab \leq 2c^2$ using the inequality above. Take the square root of both sides to get $a+b \leq c\sqrt{2}$.
A: Yet another proof uses Ptolemy's theorem. $a,b,c$ can be interpreted as sides of a triangle $ACB$, right-angled at $C$. $\triangle ACB$ is inscribed in a circle and a right-angled isosceles $\triangle ADB$ is drawn in the other semicircle as shown.

Due to Ptolemy's theorem, $AC \cdot BD + AD \cdot BC = AB \cdot CD$. Length of $CD$ can atmost be equal to that of diameter $AB$, $(CD \le c)$. Therefore
$$b\cdot \frac{c}{\sqrt{2}} + a\cdot \frac{c}{\sqrt{2}} \le c^2$$
$$\Rightarrow \boxed{a+b \le c\sqrt{2}}$$
with equality for when $CD=AB$ that is, $\triangle ABC$ is isosceles or $a=b$.
