Prove that the biggest inscribed square of an acute triangle lies on its shortest side.

Question

Assume you have an acute triangle with sides $a>b>c$. Then, after some calculation, we know that the biggest inscribed square that lies on side $a$, i.e. that has two vertices on $a$ and one vertex on each of $b$ and $c$ has side length $ah_a/(a+h_a)=2aA/(2A+a^2)$. Similarly, the biggest square that lies on side $b$ is $2bA/(2A+b^2)$ and the one that lies on $c$ is $2cA/(2A+c^2)$. I am interested in the biggest square among those three inscribed squares. I wrote a python code that tests integer values of $a, b$ and $c$. I noticed that everytime, the square that lies on $c$ is bigger than the two others, so I modified my code to count the number of cases where the square that lies on $c$ is bigger, and I got this:

counts = {'yes': 0, 'no': 0}
for a in range(1, 101):
    for b in range(1, 101):
        for c in range(1, 101):
            if a >= b >= c and a + b > c and a + c > b and b + c > a:
                S = (a + b + c) / 2
                A = (S * (S - a) * (S - b) * (S - c)) ** 0.5
                square_a = 2 * A * a / (2 * A + a ** 2)
                square_b = 2 * A * b / (2 * A + b ** 2)
                square_c = 2 * A * c / (2 * A + c ** 2)
                if square_a <= square_b <= square_c:
                    counts['yes'] += 1
                else:
                    counts['no'] += 1
print(counts)

the outcome is:

{'yes': 87125, 'no': 0}

Is there a proof that the square that lies on $c$ is always the biggest?

Answer 1

As you have already proven $l_a = \frac{2aS}{2S + a^2 }$ (where $S$ is the area of the triangle), we have: $$\begin{align} l_a \le l_b &\iff \frac{2aS}{2S + a^2 } \le \frac{2bS}{2S + b^2 }\\ &\iff\frac{a}{2S+a^2}\le \frac{b}{2S+b^2}\\ &\iff 2Sa+ab^2 \le 2Sb+ba^2\\ &\iff (2S-ab)(a-b)\le0 \tag{1} \end{align}$$ $(1)$ holds true for $a>b$ because $S = 1/2\cdot ab\sin(A)\le 1/2 \cdot ab$.

So, we have $l_a \le l_b\le l_c$ for $a>b>c$.