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

  • $\begingroup$ Since you wrote "the textbook gave no hints", I assume you took the problem from some book. Adding the name of the book to the post could be a reasonable way to provide more context. $\endgroup$ – Martin Sleziak Jun 19 '15 at 21:17

Proof without words:

enter image description here

(This space intentionally left blank.)

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Assuming you are looking for $a+b\le c\sqrt 2$, as a hint, consider $(a+b)^2+(a-b)^2$

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  • $\begingroup$ Thank you! That makes so much sense now. $\endgroup$ – anonymous Jun 19 '15 at 8:29

Let $x=(a,b)$ and $y=(1,1)$, then we have:

$a+b=x\centerdot y\leq||x||\space||y||=\sqrt{a^2+b^2}\sqrt{2}=c\sqrt{2}$.

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  • $\begingroup$ Indominus! Good to see you again and yes, I suppose your answer makes sense. Remember me from your upvoted 5 post Laying cable across a river? $\endgroup$ – anonymous Jun 19 '15 at 8:26
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    $\begingroup$ Yes, of course I remember:) But this time, I like @Blue's answer better. $\endgroup$ – Indominus Jun 19 '15 at 8:33

$(a+b)^2\\ \leq(a+b)^2+(a-b)^2\\ =a^2+2ab+b^2+a^2-2ab+b^2\\ =2a^2+2b^2 =2(a^2+b^2) =2c^2$

Using the square root gives us:


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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:

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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. $

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  • $\begingroup$ I believe you're missing an exponent on the $c$ on the second line of your proof. (It correctly appears in the next line, though.) +1. $\endgroup$ – jpmc26 Jun 19 '15 at 22:52

We solve this derivative of the Pythagorean Theorem using calculus, trigonometry, geometry, or plain old algebra, which yields the shortest, simplest proofs!


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 $a + b \leq \sqrt2 c$.


$a = c$ $\cos t$ 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.


enter image description here

$a,b,c$ 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 $2ab \leq c^2$


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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    $\begingroup$ You meant $h^2 \leq \frac{ch}{2}$. Also, please read this tutorial on how to format mathematics on this site. Using LaTeX would make your work much easier to read. $\endgroup$ – N. F. Taussig Jun 19 '15 at 11:39
  • $\begingroup$ @N.F.Taussig Yes of course, thank you. $\endgroup$ – anonymous Jun 19 '15 at 22:06

$(a-b)^2 + 2ab = c^2$. 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 $a+b \leq c\sqrt{2}$.

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