Let $ABCD$ be the square. As the circles are tangent to two sides of the square, the circles are either tangent to adjacent sides or opposite sides. If the circles touch the opposite sides of the square, the diameter is the side length of square $\implies$ circle is tangent to all sides. So all the circles are tangent to two adjacent sides of the square. Hence, the centers of the three circles must lie on either of the diagonals $AC$ and $BD$. As the circles are also tangent to each other, all three centers cannot lie on the same line (otherwise, consider two circles with centers which are farthest from each other among the three distances). Suppose centers $O_1$ and $O_2$ of circles with radius $r_1,r_2$ respectively lie on $AC$ and center $O_3$ of circle with radius $r_3$ lies on $BD$ is tangent to sides $BA$ and $BC$ (refer the figure below). We claim that $r_1=r_2$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(5.cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.0800276210439383, xmax = 5.726456218106205, ymin = -0.62759285615948, ymax = 5.68489148963604; /* image dimensions */ pair A = (0.,0.), B = (4.,0.), C = (4.,4.), D = (0.,4.), O_1 = (1.170321227001466,1.170321227001466), F = (1.170321227001466,0.), O_2 = (2.825405378560061,2.825405378560061), G = (4.,2.84494650677281), O_3 = (3.164836030317421,0.835163969682579), I = (3.164836030317421,0.), H = (4.,0.835163969682579); draw(A--B--C--D--cycle, linewidth(1.)); /* draw figures */ draw(A--B, linewidth(1.)); draw(B--C, linewidth(1.)); draw(C--D, linewidth(1.)); draw(D--A, linewidth(1.)); draw(A--C, linewidth(1.)); draw(circle(O_1, 1.170321227001466), linewidth(1.)); draw(circle(O_2, 1.1703212270014662), linewidth(1.)); draw(O_1--F, linewidth(1.)); draw(O_2--G, linewidth(1.)); draw(circle(O_3, 0.835163969682579), linewidth(1.)); draw(O_3--I, linewidth(1.)); draw(O_3--H, linewidth(1.)); draw(O_3--O_2, linewidth(1.)); draw(O_3--O_1, linewidth(1.)); /* dots and labels */ dot(A,linewidth(4.pt) + dotstyle); label("$A$", A,dir(-90)); dot(B,dotstyle); label("$B$", B, dir(0)); dot(C,dotstyle); label("$C$", C, dir(90)); dot(D,dotstyle); label("$D$", D, dir(90)); dot(O_1,dotstyle); label("$O_1$", O_1,dir(135)); dot(O_2,dotstyle); label("$O_2$", O_2, dir(135)); dot(O_3,dotstyle); label("$O_3$", O_3,dir(45)); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] WLOG $AB=1$. We have \[\sqrt{2}=AC=AO_1+O_1O_2+O_2C= r_1\sqrt{2}+(r_1+r_2)+r_2\sqrt{2}=(r_1+r_2)(1+\sqrt{2}).\]$\Rightarrow r_1+r_2=\frac{\sqrt{2}}{1+\sqrt{2}}=2-\sqrt{2}$. We also have, \[1=AB=r_1 + \sqrt{(r_1+r_3)^2-(r_1-r_3)^2}+r_3=r_1+\sqrt{4r_1r_3}+r_3.\]So, $(r_1+r_3-1)^2=4r_1r_3$. Similarly, $(r_2+r_3-1)^2=4r_2r_3$. So, \begin{align*} 0&=(r_1+r_3-1)^2-4r_1r_3\\ &=(r_1+r_3)^2+1-2(r_1+r_3)-4r_1r_3\\ &=(r_1-r_3)^2-2(r_1-r_3)+1-4r_3\\ &=(r_3-r_1+1)^2-4r_3. \end{align*}Similarly $(r_3-r_2+1)^2-4r_3=0$. So, \begin{align*} 0&=4r_3-4r_3\\ &=(r_3-r_1+1)^2-(r_3-r_2+1)^2\\ &=(r_3-r_1+1-r_3+r_2-1)(r_3-r_1+1+r_3-r_2+1)\\ &=(r_2-r_1)(2r_3+2-r_1-r_2). \end{align*}$\Rightarrow r_2-r_1=0$ or $2r_3+2-r_1-r_2=0$. If $r_1=r_2$, we are done. So suppose $2+2r_3=r_1+r_2$. So, $2r_3=r_1+r_2-2 = (2-\sqrt{2})-2=-\sqrt{2}<0$ which is not possible. Hence $r_1=r_2$ and we have proven our claim.

This is the lemma which finishes the problem quite easily.