It is easy to get $l = \max \left(\frac{ah_a}{a+h_a},\frac{bh_b}{b+h_b},\frac{ch_c}{c+h_c}\right)$ and $L = \min (\max(a,h_a),\max(b,h_b),\max(c,h_c))$. Case 1. $a =\min (\max(a,h_a),\max(b,h_b),\max(c,h_c))$. Case 1.1. $a+h_a = \min (a+h_a,b+h_b,c+h_c)$. Now $$\frac{L}{l} = \frac{a(a+h_a)}{ah_a} = \frac{a+h_a}{h_a} = \frac{a}{h_a} + 1\geq 1 + 1 =2.$$ Case 1.2. $b+h_b = \min (a+h_a,b+h_b,c+h_c)$. We have $a\geq h_a$. Suppose $b\geq h_b$ then, $a\leq b$ giving $h_a \geq h_b$. Therefore we have $b\geq a \geq h_a \geq h_b$. Also $b+h_b\leq a+h_a$ and as $ah_a = bh_b$, therefore $a=b$ and $h_a = h_b$ which is same as case 1.1. Suppose $h_b\geq b$, then $a\leq h_b$, also $a\geq h_a$. Therefore $h_b\geq a \geq h_a$, also as $bh_b = ah_a$, $h_a \geq b$. Again $h_b\geq a\geq h_a\geq b$ but $b+h_b \leq a + h_a$, so $a=h_b$ and $b = h_a$ again boiling down to case 1.1. Case 2. $h_a =\min (\max(a,h_a),\max(b,h_b),\max(c,h_c))$. Case 2.1. $a+h_a = \min (a+h_a,b+h_b,c+h_c)$. Now $$\frac{L}{l} = \frac{h_a(a+h_a)}{ah_a} = \frac{a+h_a}{a} = \frac{h_a}{a} + 1\geq 1 + 1 =2.$$ Case 2.2. $b+h_b = \min (a+h_a,b+h_b,c+h_c)$. We have $h_a \geq a$. Suppose $b\geq h_b$, then $h_a\leq b$. Then as $ah_a=bh_b$, $b\geq h_a \geq a\geq h_b$. But $b+h_b\geq a + h_a$ which gives $b=h_a$ which gives case 2.1.. If $b\leq h_b$, then $h_a \leq h_b$ and $a\geq b$. Therefore $h_b\geq h_a \geq a\geq b$ which gives $a=b$ again giving case 2.1. Equality holds for $a=h_a$ or $b=h_b$ or $c=h_c \qquad \square$ I don't know if this works

$a$ is x side of the triangle and let $h$ be the height relating to it then the maximum side of inscribed square is max $\frac{xh}{x+h}$ but the minimum side of the circumscribed square is $\underset{\text{cyclically}}{\Large\text {min}} \ \ \frac{b^2+c^2-a^2}{2\sqrt{b^2+c^2-2bc\sin A}}$...

Sketch: Let $\mathcal{S}$ and $\mathcal{S}'$ be any squares circumscribed about and inscribed in a triangle $\mathcal{T}$. The idea is to prove the easy inequalities $$2[\mathcal{S}'] \le [\mathcal{T}] \quad \text{and} \quad 2[\mathcal{T}] \le [\mathcal{S}]$$ from which the desired inequality follows. Now suppose equality holds for $\mathcal{S}$ and $\mathcal{S}'$. Letting $a$, $b$, $c$ be the sides of $\mathcal{T}$ and $h_a$, $h_b$, $h_c$ be the corresponding altitudes, the equality case for $2[\mathcal{T}] \le [\mathcal{S}]$ is $a = h_a$, $b = h_b$, or $c = h_c$, and if any of these hold it is easy to see that equality holds in $2[\mathcal{S}'] \le [\mathcal{T}]$. Thus the answer is all triangles satisfying $a = h_a$, $b = h_b$, or $c = h_c$.