This a very nice and hard problem

I wonder why this beautiful problem wasn't known until 2016.

A hard problem indeed, but I think this works. Assume that $b \ge a \ge c$. Let $I_A,I_B,I_C$ be the excenters. Then $O$, the circumcenter of $\triangle ABC$ is the nine point center of $\triangle I_AI_BI_C$, and $I$ is the orthocenter. Let $I_a$ be the distance from $I$ to $I_BI_C$; then clearly $I_a=IB$, and it is sufficient to prove that $I_a+I_b+I_c \le H_a+H_b+H_c$, where $H_b$ is the distance from $H$ to $I_AI_C$, etc. Now $A_1H \parallel OI_A$, etc. If I have time I'll post a proof of this lemma later. So $H$ lies within the angle $II_AO$, etc. Since by assumption, $b \ge a \ge c$, $H$ must lie within $AICI_B$ as well as within $II_BI_A$, which means that $H_b \ge I_b$ and $H_c \le I_c$. Since $I_a*I_BI_C+I_b*I_AI_C+I_c*I_AI_B=H_a*I_BI_C+H_b*I_AI_C+I_c*H_AI_B$, the desired result easily follows.

Here is the proof of the lemma, i.e. that $A_1H \parallel OI_A$. Let $U,V$ be the intersection of $B_1C_1$ with the circumcircle. Then $UC_1*VC_1=AC_1*BC_1=I_CC_1*C_1I$, so $I_CU,I,V$ are concyclic. So it remains to prove that if $J$ is the circumcircle of $\triangle I_BII_C$, then $O,J,I_A$ are collinear. Let $I', I_B', I_C'$ be reflections of $I,I_B,I_C$ in $O$. Then since $I,O$ are the orthocenter, nine-point center of $\triangle I_AI_BI_C$, we have that $I'$ is the circumcenter of that triangle. So $\triangle I_B'I'I_C'$ is isoceles with vertex $I'$ and $\angle I_B'I'I_C'=180- \angle BAC$. The midpoint of $I_B'I_C'$ is the midpoint of minor arc $BC$, so $\triangle I_B'I_AI_C'$ is also isoceles with vertex $I_A$ and $\angle I_B'I_AI_C'=180- \angle BAC$. Thus $J$ and $I_A$ are reflections of each other in $O$.

polya78 wrote: A hard problem indeed, but I think this works. Assume that $b \ge a \ge c$. Let $I_A,I_B,I_C$ be the excenters. Then $O$, the circumcenter of $\triangle ABC$ is the nine point center of $\triangle I_AI_BI_C$, and $I$ is the orthocenter. Let $I_a$ be the distance from $I$ to $I_BI_C$; then clearly $I_a=IB$, and it is sufficient to prove that $I_a+I_b+I_c \le H_a+H_b+H_c$, where $H_b$ is the distance from $H$ to $I_AI_C$, etc. Now $A_1H \parallel OI_A$, etc. If I have time I'll post a proof of this lemma later. So $H$ lies within the angle $II_AO$, etc. Since by assumption, $b \ge a \ge c$, $H$ must lie within $AICI_B$ as well as within $II_BI_A$, which means that $H_b \ge I_b$ and $H_c \le I_c$. Since $I_a*I_BI_C+I_b*I_AI_C+I_c*I_AI_B=H_a*I_BI_C+H_b*I_AI_C+I_c*H_AI_B$, the desired result easily follows. Could explain why does the result "easily follow"? The steps are very unclear :/

Because $(H_b-I_b)(I_AI_C-I_BI_C)+(H_c-I_c)(I_AI_B-I_BI_C) \le 0$. Add to the equality and the result falls out.

I think $H, I,$ and $F$(Fermat point of $\triangle ABC$) are collinear in this order, (I have no proof) and I think someone can prove if $X, Y, F$ are collinear in this order, $AX+BX+CX \ge AY+BY+CY \ge AF+BF+CF$, which finishes the problem. Can anyone find proof for these?

Here is a proof of the second inequality of sansae. Toss on the complex plane, and let $F = 0$, $A = a\mu$, $B = b\omega\mu$ for $b > 0$, $C = c\omega^2\mu = c > 0$, where $\omega$ is a primitive third root of unity and some $\mu$ on the unit circle. Let $x = \lambda$ for $\lambda > 0$. We will show that $f(\lambda) = AX + BX + CX$ is nondecreasing in $\lambda$. Note that \[ AX = \sqrt{|a-x|^2} = \sqrt{(a\mu-x)\left(a\frac 1{\mu} - x\right)} = \sqrt{a^2 + \lambda^2 - a\lambda k_a}, \]and similarly $BX = \sqrt{b^2+\lambda^2 - b\lambda k_b}$, $CX = \sqrt{c^2+\lambda^2 - c\lambda k_c}$, where $k_a = \mu+\frac 1{\mu}$, $k_b = \mu\omega + \frac{\omega^2}{\mu}$, and $k_c = \mu\omega^2 + \frac{\omega}{\mu}$. So, differentiating with respect to $\lambda$, we have \[ f'(\lambda) = \frac 12 \sum \frac{2\lambda - ak_a}{\sqrt{a^2+\lambda^2-a\lambda k_a}}. \]We wish to show that this is nonnegative for all $\lambda, a, b, c > 0$ ($\lambda = 0$ follows from $k_a+k_b+k_c = 0$), and $k_a, k_b, k_c$ (as $\mu$ varies). Now differentiate again with respect to $a$. We get: \[ \frac{\partial^2 f}{\partial a \partial \lambda} = \frac{\partial}{\partial a} (2\lambda - ak_a)(a^2+\lambda^2 - a\lambda k_a)^{-\frac 12} = -\frac{(2\lambda - ak_a)(2a - \lambda k_a)}{2(a^2+\lambda^2 - a\lambda k_a)^\frac 32} -k_a(a^2+\lambda^2 - a\lambda k_a)^{-\frac 12}. \]Now we solve for critical points of this. Any such point must satisfy \[ 0 = (2\lambda - ak_a)(2a - \lambda k_a) + 2k_a(a^2 - a\lambda k_a+\lambda^2) \]Fortunately we have mass cancellation and we are left with $a\lambda (4-k_a^2) = 0$, which forces $a = 0$ unless $k_a = 0$. We first solve the case where $k_a, k_b, k_c$ all have absolute value less than two. If $f_a = \frac{2\lambda - ak_a}{\sqrt{a^2+\lambda^2-a\lambda k_a}}$, we have that $f_a \geq \min(2, -k_a) = -k_a$, which are reached at $0, \infty$ respectively. Now if $k_a = 2$, then $k_b = k_c = -1$ (in particular not $\pm 2$) then \[ f_a = 2\operatorname{sgn}(\lambda-a) \geq -2 = -k_a, \]so we can still finish as before. Similarly if $k_a = -2$ then $f_a = 2 \geq -k_a$ as well. So, \[ f'(\lambda) \geq \frac 12 (-k_a-k_b-k_c) = 0 \]as desired. Remark that this fails for negative $a,b,c$ because as $a\to -\infty$, we can make $f_a = \min(\pm k_a)$ by sending $a$ to $\pm \infty$.

I do not think that $H,I,F$ are collinear... it is not mentioned in the Encyclopedia of Triangle Centers (in case anyone's interested, the three points are $X_1$, $X_{13}$ and $X_{500}$), and the points are not collinear in my Geogebra diagram.