I think geometry is not the right section. The problem is more algebra (analysis) than geometry. A general plan of proving the assertion consists of four steps: 1. Prove that $f(x)$ is non decreasing in $(0,r)$, where $r$ is the radii of the inscribed circle in $\triangle ABC$. It can be done with an appropriate choice of $DE$. 2. Prove that $f(x)$ is continuous in $(0,r)$. A proof by contradiction using (1). 3.$f(x)$ is a concave function in $(0,r)$. It folloes from the clause of the problem and (2). 4.The assertion follows from that $f$ is concave and non decreasing in $(0,r)$. Now the technical details: 1. Let assume that $0 < x_1 < x_2 < r,\, f(x_1) > f(x_2)$. Let choose $M$ from the interior of the triangle such that $d(M) = x_2$. It is easy to see that exist $D, \, E$ from the interior of the triangle such that: $M$ is midpoint of $DE$, $d(D) = d(E) = x_1$. Now we have $f(x_2) = f(d(M)) \geq \frac{1}{2}(fd(D)) + f(d(E)) = f(x_1)$ which is a contradiction. 2. First to prove that $f(x)$ is left continuous function. Assume to the contrary it doesn't hold at some point $x_0$. Then exists $\{x_i\}_{i=1}^\infty \to x_0 ,\, x_i < x_{i+1}$. We can assume that $\lim_{i \to \infty} f(x_i) = l < f(x_0)$ because it can be always achieved by choosing some subsequence of $\{x_i\}$. Choosing appropriately $x_m < x_n$, $x'=x_n+(x_n-x_m) > x_0,\, A_m, A_n, A',\, d(A_m)=x_m, d(A_n)=x_n, d(A')=x'$, we attain a contradiction with the problem's clause. Now to prove right continuity. Let assume $\{x_i\}_{i=1}^\infty \to x_0 ,\, x_{i+1} < x_{i}, \, \lim_{i \to \infty} f(x_i) = l > f(x_0) $ . Ussing left continuity of $f$ and choosing appropriately $x_n, x'=x_n-\epsilon < x_0, x''=x_n-2\epsilon$ we similarly get a contradiction. 3. From the problem clause follows: (3.1) $f(\frac{x_1+x_2}{2}) \geq \frac{1}{2}(f(x_1) + f(x_2)),\, \forall x_1,x_2 \in (0,r)$ It is well known that (3.1) and continuity of $f$ imply that $f$ is concave. Note that only (3.1) is not enough to assert $f$ is concave,but along with non decreasing of $f$ it implies concavity of $f$. 4. Denote by $d(X,AB), d(X,BC), d(X,AC)$ the distances from $X$ to $AB$, resp. $BC,\, AC$. Let $P,Q$ are interior points of $\triangle ABC$, $N \in PQ$. WLOG assume $d(N)=d(N, AB)$. From concavity of $f$ we get: $d(N)= d(N,AB) \geq \frac{QN}{PQ} d(P,AB) + \frac{PN}{PQ} d(Q,AB) \geq \frac{QN}{PQ} d(P) + \frac{PN}{PQ} d(Q) $.

dgrozev wrote: First to prove that $f(x)$ is left continuous function. Assume to the contrary it doesn't hold at some point $x_0$. Then exists $\{x_i\}_{i=1}^\infty \to x_0 ,\, x_i < x_{i+1}$. We can assume that $\lim_{i \to \infty} f(x_i) = l < f(x_0)$ because it can be always achieved by choosing some subsequence of $\{x_i\}$. Choosing appropriately $x_m < x_n$, $x'=x_n+(x_n-x_m) > x_0,\, A_m, A_n, A',\, d(A_m)=x_m, d(A_n)=x_n, d(A')=x'$, we attain a contradiction with the problem's clause. . can someone make this part a bit clear?

Please someone make this part a bit clear...Please

Cvirus213 wrote: Please someone make this part a bit clear...Please There are two sentences of the passage cited. Which one needs more explanation?