How to prove that $\frac{a+b}{c} \leq \sqrt2$, where a, b and c are the sides of a right angled triangle and c is the hypotenuse?

Question

How to prove that $\frac{a+b}{c} \leq \sqrt2$, where a, b and c are the sides of a right angled triangle and c is the hypotenuse?

My attempt:

$a+b>c$ gives $\frac{a+b}{c}$ > 1

$a<b$ and $b<c$ , adding both inequalities we get $a+b<2c$ or $\frac{a+b}{c}$<2

so for a general triangle we have the upper and lower limits of the required expression.

Now, I am confused on how to proceed further for the specific case of right angled triangle, as far as I can understand is that we need to manipulate the expression $a^2+b^2=c^2$, but can't make much progress in here

Answer 1

By AM-GM $$a^2+b^2≥2ab$$ Add $a^2+b^2$ to both sides. It's a positive value. $$2(a^2+b^2)≥a^2+b^2+2ab \implies 2c^2≥(a+b)²$$ Square root both sides and get $$\sqrt 2 c≥a+b$$

Answer 2

Consider a square with side length $a+b$ and an inner square with side length $c$ such that the inner square is incident to the outer square.

We are now going to prove the following lemma:

The smallest possible inner square area to outer square area ratio happens when the blue triangles are isosceles ($a=b$).

Proof 1 (simpler, credit: Calvin Lin):

Notice that the area of a square is equal to $\frac{1}{2}d^2$, where $d$ is the diagonal. It is relatively easy to show that the smallest that the diagonal of the inner square can be is when $d = a+b$. In all other cases, $d > a+b$ by Pythagorean.

Now it is easy to show that the smallest possible inner square will be half the area of the outer square.

$$ \frac{1}{2} (a+b)^2 \le c^2 \\\\ \frac{(a+b)^2}{c^2} \le 2 \\\\ \frac{a+b}{c} \le \sqrt{2} $$

Proof 2 (trigonometry):

As the area of a right triangle with hypotenuse $1$ is

$$\frac{\sin\theta\cos\theta}{2}=\frac{\sin 2\theta}{4}$$

then this is a maximum when $\theta=45^\circ,\;\; 0\le\theta\le45^\circ$. This implies the maximum occurs when $a=b$.

Knowing this makes it easy to calculate the area of each blue triangle (at the maximum) as $\frac{(a+b)^2}{8}$. There are $4$ such triangles, so they make up a total area of $\frac{(a+b)^2}{2}$, which is half of the outer square area (and therefore equal to the inner square area).

$$\frac{(a+b)^2}{c^2}\le2$$

Motivation:

The biggest spark of intuition comes from the $\sqrt{2}$ value, as it could have something to do with the side length ratios of a 45-45-90 degree triangle.

Playing with the values and testing some triangles might lead you to the hypothesis that the maximum probably occurs when $a=b$, further confirming that the 45-45-90 triangle is related.

Visually, this makes sense too: the "thinner" the right triangle, the closer the legs are to the hypotenuse. The isosceles right triangle pulls the legs furthest away from the hypotenuse.

Once the goal is more clear, now it is just a matter of how to prove such an idea. Here, it helps to see that when the maximum of $\frac{a+b}{c}$ is attained, the area of the right triangle is also maximized (relative to fixing $a+b$ or $c$). Combining that with some basic algebra rearrangement could therefore lead to this proof to make these intuitions rigorous.

Answer 3

Using $a^2+b^2=c^2$ we have $$ 0\leq(a-b)^2 \implies 2ab\leq c^2 \\ \implies a^2+b^2+2ab\leq 2c^2 \implies a+b\leq \sqrt{2}c. $$

Simple version of the solution by @Oppenheimer.

Answer 4

As you have pointed out,

a + b $>$ c

to prove your question,

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

square both the sides,

$a^{2}$ + $b^{2}$ + 2ab $\leq$ $2c^{2}$

since $a^{2} + b^{2} = c^{2}$,

it becomes,

$2ab$ $\leq$ $c^{2}$

Case - I, suppose $2ab = c^{2}$

substituting, $c^{2} = a^{2} + b^{2}$,

$( a - b )^{2} = 0$

this case is true when a = b,

Case - II, $2ab$ $\leq$ $c^{2}$ We shall prove this result by way of contradiction,

Assume the opposite,

$2ab$ $\geq$ $c^{2}$

suppose 2ab - k = $a^{2} + b^{2}$

where k is a positive integer,

-k = $( a - b )^{2}$

Which cannot be true, as the square of anything is positive ( Ignoring complex numbers )

Which completes our proof.