Solution:
We start with a lemma.

Lemma. Let $R$ be the circumradius of a quadrilateral $X Y Z T$, the diagonals of $X Y Z T$ meet at $U$, and $\varphi=\frac{1}{2} \angle X U Y$. Then the radii $r_{1}$ and $r_{2}$ of the incentres of $X Y U$ and $Z T U$ satisfy
$$\frac{r_{1}+r_{2}}{R} \leqslant 2 \tan \varphi(1-\sin \varphi)$$
Indeed, let $\angle U X Y=2 \psi$, $\angle U Y X=2 \vartheta$, then $\angle U T Z=\angle U X Y=2 \psi$, $\angle U Z T=\angle U Y X=2 \vartheta$ (and obviously $\psi+\vartheta+\varphi=\frac{\pi}{2}$). We have $X Y+Z T=\left(r_{1}+r_{2}\right)(\cot \psi+\cot \vartheta)=2 R \sin \angle X T Y+2 R \sin (2 \varphi-\angle X T Y)=2 R(\sin \angle X T Y+\sin (2 \varphi-\angle X T Y))=2 R \cdot 2 \sin \varphi \cos (\varphi-\angle X T Y) \leqslant 4 R \sin \varphi$. Therefore
$$\begin{gathered} \frac{r_{1}+r_{2}}{R} \leqslant \frac{4 \sin \varphi}{\cot \psi+\cot \vartheta}=\frac{4 \sin \varphi \sin \psi \sin \vartheta}{\sin (\psi+\vartheta)}=\frac{4 \sin \varphi \sin \psi \sin \vartheta}{\cos \varphi}=4 \tan \varphi \sin \psi \sin \vartheta= \\ =4 \tan \varphi \cdot \frac{1}{2}(\cos (\psi-\vartheta)-\cos (\psi+\vartheta)) \leqslant 2 \tan \varphi(1-\sin \varphi) \end{gathered}$$
q.e.d.

Returning to the problem, let $\angle A M B=2 \alpha$, $\angle B N C=2 \beta$, $\angle C K D=2 \gamma$, then $\alpha+\beta+\gamma=\frac{\pi}{2}$.

Applying the inequality (1) to the quadrilaterals $A B D E$, $B C E F$ and $C D F A$ we get
$$\frac{r_{1}+r_{2}+r_{3}+r_{4}+r_{5}+r_{6}}{R}=\frac{r_{1}+r_{4}}{R}+\frac{r_{2}+r_{5}}{R}+\frac{r_{3}+r_{6}}{R} \leqslant 2 \tan \alpha(1-\sin \alpha)+2 \tan \beta(1-\sin \beta)+2 \tan \gamma(1-\sin \gamma)$$
We claim that if $\alpha+\beta+\gamma=\frac{\pi}{2}$ then
$$2 \tan \alpha(1-\sin \alpha)+2 \tan \beta(1-\sin \beta)+2 \tan \gamma(1-\sin \gamma) \leqslant \sqrt{3}$$
To prove that we consider the function $f(x)=2 \tan x(1-\sin x)$ for $x \in\left(0 ; \frac{\pi}{2}\right)$.
Since $f''(x)=-2 \frac{(1-\sin x)^{2}+\cos ^{4} x}{\cos ^{3} x}<0$ for $x \in\left(0 ; \frac{\pi}{2}\right)$, it follows from Jensen's inequality that
$$f(\alpha)+f(\beta)+f(\gamma) \leqslant 3 f\left(\frac{\alpha+\beta+\gamma}{3}\right)=3 f\left(\frac{\pi}{6}\right)=\sqrt{3}$$
Thus (2) is proved, and $r_{1}+r_{2}+r_{3}+r_{4}+r_{5}+r_{6} \leqslant \sqrt{3} R$.