Since $\frac{\overline{AM}}{\overline{MB}}\times\frac{\overline{BK}}{\overline{KC}}\times\frac{\overline{CL}}{\overline{LA}}=1$ WLOG let $\frac{\overline{AM}}{\overline{MB}}\ge1$ and $\frac{\overline{BK}}{\overline{KC}}\le1$ if $\frac{\overline{CL}}{\overline{LA}}\ge1$ then $\frac{\overline{KC}}{\overline{BK}}\ge\frac{\overline{CL}}{\overline{LA}}$ and $\frac{\overline{AM}}{\overline{MB}}\ge\frac{\overline{AL}}{\overline{LC}}$ Let the radius of the incircle of triangle $AML$, $CKL$, $ABC$ be $r_1$, $r_2$, $r$, respectively then we have $r_1+r_2\ge\frac{\overline{AL}}{\overline{AC}}r+\frac{\overline{CL}}{\overline{AC}}r=r$ if $\frac{\overline{CL}}{\overline{LA}}\le1$ then $\frac{\overline{AM}}{\overline{MB}}\ge\frac{\overline{AL}}{\overline{LC}}$ and $\frac{\overline{CK}}{\overline{KB}}\ge\frac{\overline{CL}}{\overline{LA}}$ similar to the case $\frac{\overline{CL}}{\overline{LA}}\ge1$, we get the same result. Done!

wanwan4343 wrote: $r_1+r_2\ge\frac{\overline{AL}}{\overline{AC}}r+\frac{\overline{CL}}{\overline{AC}}r=r$ Can you explain how??

Why is $\frac{\overline{KC}}{\overline{BK}}\ge\frac{\overline{CL}}{\overline{LA}}$? in the case $\frac{\overline{CL}}{\overline{LA}}\ge1$?

Anyone have an explanation? Thanks!

Since no one else will, I shall do so: Let $L$, $M$, and $K$ are in the configuration shown, and $Q$ is the point on $AC$ such that $MQ \parallel BC$ and $P$ be the point on $BC$ such that $MP \parallel AC$. Also define the function $r(ABC)$ to be the inradius of triangle $ABC$. Claim: $r(AML) > r(AMQ)$ Proof: This is the intuitive part of the proof. The incircle of $AML$, as well as the incircle of $AMQ$, is tangent to $AM$ and $AQ$. However, we need to "extend" the circle from being tangent to $MQ$ to being tangent to $ML$. The rest of the problem is not bad at all. If $L$ and $K$ are in the specified configurations, we will be able to conclude that \[ r(AML) + r(BMK) \ge r(AMQ) + r(BMP) = \frac{AM}{AB}r(ABC) + \frac{MB}{AB}r(ABC) = r(ABC) \], as desired. It remains to show that we may always find $L$ and $K$ in such configurations. We will show that this argument must work for $\textit{some}$ pair of triangles, which is the assertion. Now let the ratios $\frac{BM}{MA} = x$, $\frac{AL}{LC} = y$, $\frac{CK}{KB} = z$. Observe that $L$ below $Q$ condition is equivalent to $xy < 1$, and that the $K$ after $P$ condition is equivalent to $xz > 1$. By Menelaus's theorem, this is equivalent to $y < 1$ and $z > 1$, but it is easy to see that there must be two sides with this property, since $xyz > 1$. Hence, we are done.

I was trying to do this problem by proving that $r_1+r_2+r_3\geq{\frac{3}{2}r}$, is this a viable strategy?

fprosk wrote: I was trying to do this problem by proving that $r_1+r_2+r_3\geq{\frac{3}{2}r}$, is this a viable strategy? Nope, this cannot work. Think about what happens if one of the three triangles takes up most of $ABC$.

Let $x=\frac{AM}{MB},y=\frac{BK}{KC}, z=\frac{CL}{LA}$. There exists a permutation$\{ a,b,c \}$ of $\{x,y,z \}$ such that $ab \geq 1$ and $ac \leq 1$. Assume $xy \geq 1$. Then AFSoC, $xz>1$ and $yz>1$. Then $(xyz)^2>1$, contradicting Ceva ($xyz=1$). If $xy \leq 1$ we can proceed analogously. Now, WLOG $xy \geq 1$ and $xz \leq 1$. Consider the inradii $r_1,r_2$ of triangles $AML,BMK$. we have $\frac{BM}{MA} \leq \frac{BK}{KC}$ and $\frac{AM}{MB} \leq \frac{AL}{LC}$. Then let $R,S$ be points on sides $AC,AB$ respectively such that $MR \parallel AB$ and $MS \parallel AC$. Clearly $r_{AMR} \leq r_1$ and $r_{BMS} \leq r_2$. But, $$\frac{r_1}{r} +\frac{r_2}{r} \geq \frac{r_{AMR}}{r} + \frac{r_{BMS}}{r} = \frac{AM}{AB} + \frac{BM}{AB}=1$$So we are done.

For arbitrary triangle $XYZ$ we denote it's inradius as $r(XYZ).$ By Ceva $\frac{|BK|\cdot |CL|\cdot |AM|}{|KC|\cdot |LA|\cdot |MB|}=1$ so WLOG $$|BK|:|KC|\geq 1,|CL|:|LA|\leq 1\implies |CK|:|KB|\leq |AM|:|MB|\leq |AL|:|LC|.$$Now we may construct points $L',K'$ on segments $AL,BK$ such that $ML'\parallel BC,MK'\parallel AC$. Finally we obtain $$r(AML)+r(BMK)\geq r(AML')+r(BMK')=\frac{|AM|}{|BC|}r(ABC)+\frac{|BM|}{|BC|}r(ABC)=r(ABC).$$

Beautiful problem. Here's a rough sketch (I omitted a few details). Suppose otherwise. Draw lines parallel to $AC,AB$ which pass through $K$ and intersect $AB,AC$ at $K_B$ and $K_C$. Note that we can't have triangles $BKM$ and $CKL$ contain $BKK_B$ and $CKK_C$ at the same time. However, we also can't have both triangles not contain $BKK_B$ and $CKK_C$ since if we draw lines parallel to $BC$ passing through $M,L$ which intersect $AC,AB$ at $M_A$ and $L_A$ we see that at least one of $AMM_A$ and $ALL_A$ contains $AML$. Then we see that we must have a configuration in which either $\frac{AM}{AL},\frac{BK}{BM},\frac{CL}{CK}$ are greater than $\frac{AB}{AC},\frac{BC}{AB},\frac{AC}{BC}$, or vice versa, meaning that the lines don't concur by Ceva's Theorem, a contradiction. $\blacksquare$

Equality case kills (medial triangle, $LM, MK, KL$ are parallel $BC, CA, AB$). Draw the line $\ell_A$ parallel to $BC$ and tangent to the incircle of $ALM$ closer to $BC$, similarly construct $\ell_B, \ell_C$, then by Ceva and the cyclic product of the value $\frac{AL\cdot AB}{AC \cdot AM}$, the value cannot be $>1$ (resp $<1$) for all cyclic variation, hence there exists a configuration such that WLOG $\ell_A, \ell_B$ intersect $AB$ at $P_A, P_B$ such that $A, P_B, M, P_A, B$ appear on $AB$ in that order, so $\ell_A$ intersects $\ell_B$ inside $\triangle ABC$, so by homothety at $A, B$ we are done by choosing $ALM, BMK$.

Note that by Ceva's theorem, we have \[\left(\frac{BL}{LA}\cdot \frac{AK}{CK}\right)\left(\frac{AK}{KC}\cdot \frac{CM}{MB}\right)\left(\frac{CM}{MB}\cdot \frac{BL}{LA}\right)=1.\]Therefore, one of those factors must be at least one, and one must be at most one. Without loss of generality, let \begin{align*} \frac{BL}{LA}\cdot \frac{AK}{CK} &\le 1 \\ \frac{CM}{MB}\cdot \frac{BL}{LA} & \ge 1 \end{align*}Then what we have is that $\frac{BL}{LA}\le \frac{CK}{AK}.$ Thus, if we draw the line through $K$ parallel to $BC$ meeting $AB$ at $X$, then $X$ lies between $A$ and $L$ inclusive. Similarly, if we draw the line through $K$ parallel to $AB$ meeting $BC$ at $Y$ then $Y$ lies between $M$ and $C$ inclusive. $~$ Now, since $\triangle AXK$ is completely inside of $\triangle ALK$, and the incircle is the largest circle that can be placed completely inside a triangle, the incircle of $\triangle AXK$ cannot be larger than that of $\triangle ALK$, and similarly the incircle of $\triangle YKC$ cannot be larger than that of $\triangle MKC.$ Therefore, the sum of the inradii of $\triangle ALK$ and $\triangle MKC$ is at least the sum of the inradii of $\triangle AXK$ and $\triangle YKC$, which is equal to the inradius of $\triangle ABC$, and so we're done.

I'm surprised how nice I found this given how strange the problem statement looks at first glance. This solution is motivated by the equality case of the medial triangle (particularly, parallel lines) $\textbf{Claim.}$ We have that at least one of the statements \[ \frac{BM}{MA} \le 1 \wedge \frac{BK}{KC} \le 1 \]and cyclic is true. $\textit{Proof.}$ This follows from Ceva; we have \[ \frac{BM}{MA} \cdot \frac{AL}{LC} \cdot \frac{CK}{KB} = 1, \]and more particularly, WLOG assuming $\frac{BM}{MA} \le 1$, we conclude $\frac{BK}{KC} > 1$ and $\frac{AL}{LC} > 1$ which yields a true result for $C$. We now let $A$ be the vertex which satisfies the above property. Remark that \begin{align*} \frac{AM}{BM} &\le 1 \\ \iff 1 &\le \frac{BK}{CK} \cdot \frac{CL}{AL} \\ \iff \frac{AL}{CL} &\le \frac{BK}{CK} \\ \iff \frac{AC}{CL} &\le \frac{BC}{CK} \\ \iff \frac{CL}{AC} &\ge \frac{CK}{BC}, \end{align*}which means that $KL$ intersects $AB$ closer to $A$, and more particularly, the line through $K$ parallel to $AB$ intersects the interior of segment $CL$. Thus, denoting the inradius of $\triangle CKL$ as $r_C$, \[ r_C \ge \frac{CK}{BC} \cdot r. \]In exactly the same manner, denoting the inradius of $\triangle BKM$ as $r_B$, \[ r_B \ge \frac{BK}{BC} \cdot r.\]Summing yields $r_C + r_B \ge r$, as desired.