# Prove that the area of a right triangle is no more than the square of the hypotenuse divided by 4

I'm struggling with this proof. Let $$a,b,c$$ denote the base, height, and hypotenuse of a right triangle. Let $$A$$ denote area of this triangle

We have the following relations: $$a^2 + b^2 = c^2 \\ A = a*b/2$$

and also the triangle inequality, though I'm not sure if this inequality comes into play.

Here is some of my work and I'm unsure if it's in the right direction $$a^2 + b^2 = c^2 \\ a^2*b + b^3 = c^2*b \\ \implies A*a + b^3 = b * c^2 \\ \implies A = b*c^2/a - b^3/a$$

I feel that this is not in the right direction as I don't know how to show that this last expression is $$\leq c^2/4$$. Can anyone advise?

• you need to prove $a^2 + b^2 \geq 2ab$. Feb 25, 2021 at 17:16
• @MathLover Ah I didn't start from that. I'll look into this, but it seems there's some property that would make proving this relationship relatively simple? Feb 25, 2021 at 17:19
• $(a-b)^2$ has to be $\geq 0$, right? That is all there is to the proof. Feb 25, 2021 at 17:21
• @MathLover Yeah. Just realized that. I wasn't thinking about completing the square initially. Feb 25, 2021 at 17:22

The claim is $$(a^2+b^2)/2\ge ab$$, which is the AM-GM inequality in $$2$$ positive variables. In particular, $$a^2+b^2-2ab=(a-b)^2\ge0$$.

• Ah that was super simple.. My mind gravitated to not starting with the claim, but starting from the definition of area and the pythagorean theorem.. Feb 25, 2021 at 17:20

Here is a picture to visualize the property.

Any right triangle lies within the semicircle erected on a square with side length equal to the hypotenuse.

All such triangles have height less than or equal to the radius of semicircle i.e., half the hypotenuse length. The right triangle with maximum area can be seen as covering exactly one-fourth of the above square. Hence, $$\text{Area of any right triangle} \le \frac{(\text{Hypotenuse})^2}{4}$$

The maximum (equality) is achieved by right isosceles triangles.

Well, the area the triangle is $$\frac 12ab$$ and the hypotenuse is $$\sqrt{a^2 + b^2}$$ and we need to show $$\frac 12 ab \le \frac {a^2 + b^2}4$$.

This is equivalent to proving $$ab \le \frac {a^2 + b^2}4$$.

Are you familiar with the AM-GM inequality?

If you have two positive (or non-negative... but if either is $$0$$ it is trivial) $$m,n$$ then the Arithmetic Mean, $$\frac {m+n}2$$, is greater or equal to the Geometric Mean, $$\sqrt{mn}$$.

This is exactly that. Let $$m = a^2 > 0$$ and $$n = b^2 > 0$$ then $$\frac {m+n}2 \ge \sqrt{mn}$$ so $$\frac {a^2 + b^2}2 \ge ab$$ and $$\frac {a^2 + b^2}4 \ge \frac 12 ab$$.

That's that.

....

Unless you aren't familiar with AM-GM theorem.

Pf: Note that $$(a-b)^2 \ge 0$$ with equality holding only if $$a = b$$.

So $$a^2 -2ab +b^2 \ge 0$$ and so $$a^2 + b^2 \ge 2ab$$ and $$\frac {a^2 + b^2}2 \ge ab$$.

Replace $$m =\sqrt a$$ and $$n= \sqrt b$$ (assuming $$a,b > 0$$) to get the AM-GM inequality.

$$\frac {m+n}2$$ is the Arithemetic Mean (or average). https://en.wikipedia.org/wiki/Arithmetic_mean

And for $$m,n \ge 0$$ we have $$\sqrt{mn}$$ is the Geometric Mean https://en.wikipedia.org/wiki/Geometric_mean

The way to think of the geometric mean to imagine you have a rectangle with sides $$a,b$$ and area $$ab$$ and you want to find the "average" of the sides so that the area stays the same but the sides are "typical" and about the same. That is the "average" of $$a,b$$ is $$s$$ where $$s\times s = ab$$ or $$s = \sqrt{ab}$$.

That's actually the same philosophy as the arithemetic mean. If we have $$a + b = T$$ some sum and we want to "average" the terms so that the sum $$T$$ stays the same but the terms are "typical" and about the same, we want the "average" of $$a,b$$ to be $$m$$ where $$m + m = a+b$$ or $$m = \frac {a+b}2$$.

The only difference is that for geometric mean we want a product to stay the same while the factors "average out". ANd for the arithmetic mean we want a sum to stay the same while the addends "average out".

......

Another way to grok why the AM-GM theorem should feel right-- If I look and $$\frac {a+b}2 \ge \sqrt ab$$, I don't see anything that makes me think "well, duh, obvious". In fact I think, well why the heck should that be true.

But.... if we consider that for any $$a,b$$ we can consider that for if we assume $$a < b$$ then there is a midpoint $$m$$ between them and $$a = m-d$$ and $$b= m+d$$ where $$d$$ is have the distance between $$a$$ and $$d$$.

Then $$ab = (m-d)(m+d) = m^2 - d^2$$ and of course $$m^2 - d^2 < m^2$$.

So $$\sqrt {ab} < m = \frac {a+b}2$$.

Well, duh, obvious. .... (no... not really).

It is the same thing as saying that among all right triangles sharing a common hypotenuse the right isosceles triangle has maximum area:

If $$a^2+b^2$$ is a constraint, maximum area $$ab$$ occurs when $$a=b$$, can be established by plugging either $$a$$ or $$b$$ into area $$ab$$ and differentiating...

The area of all other right triangles come out less than the gray triangle shown.