Suppose $a,b,c$ are lengths of the sides of the triangle $ABC$, $h_a,h_b,h_c$ are the lengths of the respective altitudes and $s_a,s_b,s_c$ are sides of the respective squares. It is easy to get $s_a=\frac{ah_a}{a+h_a}$ from similarity. We need to show $s_a^2+s_b^2>s_c^2$ and its given that $a^2+b^2>c^2$ and so on. So we need \[\left(\frac{ah_a}{a+h_a}\right)^2+\left(\frac{bh_b}{b+h_b}\right)^2>\left(\frac{ch_c}{c+h_c}\right)^2\] Note that $ah_a=2[ABC]$ where $[ABC]$ is the area of $\triangle ABC$. So we need $\left(\frac{1}{a+h_a}\right)^2+\left(\frac{1}{b+h_b}\right)^2>\left(\frac{1}{c+h_c}\right)^2$. Let $a^2=x,b^2=y,c^2=z,k=2[ABC]$. As $ABC$ is acute angled, $x+y>z$ and so on. From heron's formula, \[k^2=(a+b+c)(a+b-c)(a-b+c)(-a+b+c)=2(a^2b^2+b^2c^2+c^2a^2)-a^4-b^4-c^4=2xy+2yz+2zx-x^2-y^2-z^2.\] We have \begin{align*} (c+h_c)^2(a+h_a)^2&=\left(c+\frac{k}{c}\right)^2\left(a+\frac{k}{a}\right)^2\\ &= \left(c^2+2k+\frac{k^2}{c^2}\right)\left(a^2+2k+\frac{k^2}{a^2}\right)\\ &= \left(z+2k+\frac{k^2}{z}\right)\left(x+2k+\frac{k^2}{x}\right)\\ &= xz+\frac{k^4}{xz}+k^2\left(\frac{x}{z}+\frac{z}{x}\right)+4k^2+2k\left(x+z+\frac{k^2}{z}+\frac{k^2}{x}\right). \end{align*} So we need $(xz+yz-xy)+k^4\left(\frac{1}{xz}+\frac{1}{yz}-\frac{1}{xy}\right) +k^2\left(\frac{x}{z}+\frac{z}{x}+\frac{y}{z}+\frac{z}{y}-\frac{x}{y}-\frac{y}{x}\right) +(4k^2+4k^2-4k^2)+2k\left((x+z+y+z-x-y)+\left(\frac{k^2}{z}+\frac{k^2}{x}+\frac{k^2}{y}+\frac{k^2}{z}-\frac{k^2}{x}-\frac{k^2}{y}\right)\right)>0.$ Post simplifying we need to show \[(xz+yz-xy)+\frac{k^4}{xyz}(x+y-z) +k^2\left(\frac{x}{z}+\frac{z}{x}+\frac{y}{z}+\frac{z}{y}-\frac{x}{y}-\frac{y}{x}\right) +4k^2+ 2k\left(2z+\frac{2k^2}{z}\right)>0.\] We have $x+y-z>0$. $k^2\left(\frac{x}{z}+\frac{z}{x}+\frac{y}{z}+\frac{z}{y}-\frac{x}{y}-\frac{y}{x}\right) +4k^2=k^2\left(\frac{x+y}{z}+\frac{z-x}{y}+\frac{z-y}{x}+4\right)>k^2\left(\frac{z}{z}+\frac{-y}{y}+\frac{-x}{x}+4\right)=3k^2>0$. So we need to show $xz+yz-xy+3k^2+2k\left(2z+\frac{2k^2}{z}\right)>0$.

Let $a, b, c$ be the sides of the triangles, $x, y, z$ be the sides of the square, $d$ be the circumdiameter,, $h_a, h_b, h_c$ be the altitudes. Claim 1:$x = \frac{abc}{ad + bc}$ and cyclic variations. [asy][asy]unitsize(2cm); pair A = dir(108), B = dir(210), C = -1/B; pair C_1 = rotate(90, C)*B; pair B_1 = rotate(-90, B)*C; pair X_1 = intersectionpoint(A--B_1, B--C); pair X_2 = intersectionpoint(A--C_1, B--C); pair X_3 = rotate(-90,X_2)*X_1; pair X_4 = rotate(90,X_1)*X_2; draw(A--B--C--cycle); draw(B--B_1--C_1--C, dotted); draw(X_1--X_4--X_3--X_2); draw(B_1--A--C_1, linetype("0 2")); dot(A);dot(B);dot(C); dot(X_1);dot(X_2); dot(C_1); dot(B_1); label("A", A, dir(90)); label("B", B, dir(180)); label("C", C, dir(0)); label("X$_1$", X_1, dir(135)); label("B$_1$", B_1, -dir(45)); label("C$_1$", C_1, -dir(135)); label("X$_2$", X_2, dir(45));[/asy][/asy] Consider the homothety mapping the square such that two of its vertices is on $BC$ to the square on side $BC$ lying outside of the triangle. \[\frac{x}{h_a} = \frac{a}{a+h_a} \iff x = \frac{ah_a}{a + h_a} = \frac{bc\sin A}{a + \frac{bc}{a}\sin A} = \frac{abc}{\frac{a^2}{\sin A} + bc} = \frac{abc}{ad + bc}\] Claim 2 $a\ge b\ge c$ implies that $x\ge y\ge z$. Note that $d > a, b, c$. \[(a - b)d \ge (a - b)c \iff ad + bc \ge bd + ca \iff \frac{1}{bd + ca} > \frac{1}{ad + bc} \iff \frac{abc}{bd + ca} > \frac{abc}{ad + bc} \iff y \ge x\]\[(b - c)d \ge (b - c)a \iff bd + ca \ge cd + ab \iff \frac{1}{cd + ab} \ge \frac{1}{bd + ca} \iff \frac{abc}{cd + ab} \ge \frac{abc}{bd + ca} \iff z \ge y\] We can assume without loss of generality that $a\ge b\ge c$. Note that $a^2 \ge ab \ge ca, b^2 \ge bc \ge c^2$ and $b^2 + c^2 > a^2$ and cylic variations. Thus, it suffices to prove that $x + y > z$ and $x^2 + y^2 > z^2$. However, the first follows from the second since $(x + y)^2 > x^2 + y^2 > z^2$. Thus it suffices to prove that \begin{align*} x^2 + y^2 &> z^2\\ \iff (cd + ab)((ad + bc)^2 + (bd + ca)^2) &> (ad + bc)^2(bd + ca)^2 \end{align*}