Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

This is one problem I had in one TST in 2000. Solution: Consider the following lemma: let $ABC$ be a triangle and let $M$ be a point on the side $AC$ such that $\angle BMC \geq 90^\circ$. Then the following inequality takes place: \[ BM+MC < BA + BC \]

Back to the problem, we notice that at least two of the angles $\angle AMB$, $\angle BMC$, $\angle CMA$ are larger than $90^\circ$. Indeed, if two of them are less than $90^\circ$, since their sum is $360^\circ$, one of them would be larger than $180^\circ$ which is an obvious contradiction. So, let us suppose WLOG that $\angle AMB \geq 90^\circ$ and $\angle AMC \geq 90^\circ$. Also consider $M' = AM \cap BC$. We apply the lemma for the triangles $ABM'$ and $ACM'$ and we obtain \[ AM + MB < AB + BM' \eqno (1) \] \[ AM + MC < AC + CM' \eqno (2) \] From (1) and (2), by addition, we obtain \[ AM+BM+ CM + AM < AB + BC + CA \] which solves our problem. Also there is the official solution, much complicated which I will post later.

That's a beautiful solution! Just a little remark about your proof of the lemma: Valentin Vornicu wrote: Consider the following lemma: let $ABC$ be a triangle and let $M$ be a point on the side $AC$ such that $\angle BMC \geq 90^\circ$. Then the following inequality takes place: \[ BM+MC < BA + BC \]

In this proof, you needn't distinguish between the cases $\angle BAC \leq 90^\circ$ and $\angle BAC > 90^\circ$. In both cases, your argument works:

Darij

Indeed

Valentin Vornicu wrote: Also there is the official solution, much complicated which I will post later. I don't know what you had mind but these solutions are from the IMO1999 booklet I typed once. I always pasted them on ML as solutions, usually by grobber, were given. Lemma If $M$ is an interior point of the convex quadrilateral $ABCD$ then $MA+MB<AC+CD+DB$. Proof of the Lemma The ray $AM$ intersects the quadrilateral in $N$; suppose, for instance, that $N \in [CD]$. Then \[MA + MB < MA + MN + NB \leq AN + NC + CB \leq AD + DN + NC + CB = AD + DC + CB.\] Offical Solution: The median triangle $ D E F $ divides triangle $ ABC $ into four regions. Each region is covered by at least two of the convex quadrilaterals $ABDE, BCEF, CAFD$. If, for instance, $M$ belongs to $[ABDE]$ and $[BCEF]$ then $MA + MB < BD + DE + EA$ and $MB + MC < CE + EF + FB$. By adding these two inequalities we get $MB+(MA+MB+MC) \leq AB + BC + CA$, which implies the required conclusion.

Another Offical Solution: Let $O$ be the circumcenter of $\triangle ABC$. - Case 1: $O \in [ABC]$ . Suppose $M \in [ADOE]$. Then $MA = min\{MA,MB,MC\}$ and $MA +MB \leq NA + NB, MA+MC \leq NA + NC$. Since $NA + NC \leq AD + DC$. (for $O([ANC]$; otherwise $NA + NC < BD + DC$) it is enough to prove that (with the usual notations). Since $m_{c} < (a+b)/2$, it is enough $m_{c} + (c/2) + 2R < a+b+c$ to prove that $4R < a+b+c$ (in an nonobtuse-angled triangle). This reduces to $\sin A+\sin B+\sin C >2$, or \[ \sin \frac{A}{2}\cos \frac{A}{2}+ \cos \frac{A}{2} \cos \frac{B-C}{2} \frac{}{}>1. \] For fixed A the minimum value of the left member is obtained when $B-C$ has maximum value, that is $B= \pi/2, C+A =\pi/2$ and the minimum is $(1/2)(\sin A + 1 + \cos A)$ therefore it is greater than $(1/2)(1+1)=1$ - Case 2: $ABC$ is obtuse-angled. Suppose for instance that $m(\widehat{B})>90^{\circ}$. Let $P$ and $Q$ be the intersections of the perpendicular bisectors of the sides $[AB]$ and $[BC]$ with $AC$. Case 2.1: For $M \in [ADP], min\{MA,MB,MC\}=MA$ and, as above, $MA + MB \leq PA + PB = 2 \cdot AP, MA + MC \leq m_{c} + (c/2) < (a + b + c)/2$. It is enough to prove that $4 \cdot AP < a+b+c$, which is implied by $4AP < 2b < a+b+c$. A similar argument works in the case $M \in [CEQ]$. Case 2.2: For $M \in [BEQPD]$, let $BM \cap AC = N$. Then $min\{MA,MB,MC\}= MB$ and $MB \cdot MC \leq NB + NC, MB+MA \leq NB + NA, NB \leq max\{PB,QB\}$, therefore it is enough to prove that $2 \cdot PB < AB + BC$. Since $PB < BD + DP < BD + PP'$, the needed relation is obvious.

We can also prove that \[ 2(\min \{MA,MB,MC\} + \max \{MA,MB,MC\})\le AB + BC + CA \] Equality holds only if the triangle is degenerate (but not iff). Proof: Lemma: Given two lengths $ a\ge b$, and a point $ M$, let $ XYZ$ be the triangle (possibly degenerate) with smallest perimeter such that $ MX = a$ and $ MY = MZ = b$ and $ M$ lies in the interior of $ XYZ$ or on it's sides (If there are more than one then let one of them $ XYZ$) We shall prove that the perimeter of $ XYZ\ge 2(a + b)$ Denote the circle with radius $ b$ and center $ M$ to be $ (C)$ Denote the ellipse with foci $ X,Y$ passing through $ Z$ to be $ (E)$ (If the ellipse is degenerate then $ X,Y,Z$ collinear and easy to see that $ 2(a + b)\le XY + YZ + ZX$, so we assume the contrary) $ (C)$ and $ (E)$ has (at least) one common point $ B$. If they ($ (C)$ and $ (E)$) are not tangent to each other then there also exists a point $ Z'$ which lies on the circumference of $ (C)$ and in the interior of $ (E)$ Hence $ YZ + XZ > YZ' + XZ'$ Then The triangle $ XYZ'$ satisfies $ MX = a$ and $ MY = MZ' = b$ but the perimeter of $ XYZ'$ is less than that of $ XYZ$. Contradiction. Hence $ (C)$ and $ (E)$ are tangent to each other. By the well known reflective property of ellipse $ \angle XZM = \angle YZM$. By symmetry $ \angle XYM = \angle ZYM$ Hence $ M$ is the incenter of $ XYZ$. Also $ MY = MZ$ implies $ XYZ$ is isosceles. Let $ XM$ extended meet $ YZ$ at $ T$. $ XT$ is also median, altitude of triangle $ XYZ$. Let $ YT = ZT = x$ and $ \angle MYT = \angle MZT = \cdots = \theta$ We can obtain that \[ YZ = 2x, MT = x\tan \theta , XT = x\tan 2\theta , a = XM = x\tan 2\theta - x\tan \theta \] \[ b = MY = MZ = x\sec \theta , XY = XZ = x\sec 2\theta \] So: \[ 2(a + b)\le XY + XZ + YZ \] \[ \iff 2[(x\tan 2\theta - x\tan \theta ) + (x\sec \theta )]\le (x\tan \theta ) + (x\tan \theta ) + 2x \] \[ \iff 1 + \tan \theta + \sec 2\theta \ge \tan 2\theta + \sec \theta \] \[ \iff 1 + \frac {1}{\tan 2\theta + \sec 2\theta }\ge \frac {1}{\tan \theta + \sec \theta } \] which is true since $ \theta \in (0,\frac {\pi}{4})$. End Lemma Now let $ ABC$ any arbitrary triangle. (unlike $ XYZ$). WLOG $ MA = \max \{MA,MB,MC\}$ and $ MB = \min \{MA,MB,MC\}$ Cut $ MC'$ equal to $ MB$ from $ MC$. Perimeter of $ ABC\ge$ that of $ ABC'$ From lemma with $ a = MA$ and $ b = MB = MC'$, we have perimeter of $ ABC'\ge 2(MA + MB)$ Hence perimeter of $ ABC\ge 2(MA + MB)$ QED

We will show that for all $M$ not in the exterior of $\triangle ABC$, $\min(2MA + MB + MC, MA + 2MB + MC, MA + MB + 2MC\) < AB + BC + CA$.For any $X$ (not necessarily in the interior of $\triangle ABC$), let $f_1(X) = 2AX + BX + CX$, $f_2(X) = AX + 2BX + CX$, $f_3(X) = AX + BX + 2CX$, and $f(X) = \min(f_1(X), f_2(X), f_3(X))$. We wish to show that $f(X) < AB + BC + CA$. We will first prove three lemmas: Lemma 1: If $O$ is the circumcenter of acute $\triangle ABC$, $f(O) < AB + BC + CA$. Proof: Let $R$ be the circumradius of $\triangle ABC$. $f(O) = 4R$, so we simply wish to show that $\frac{AB}{2R} + \frac{BC}{2R} + \frac{CA}{2R} > 2$. Dividing by $2R$ and using the law of sines, we find that the inequality is equivalent to $\sin A + \sin B + \sin C > 2$. Because $\frac{d^2 \sin x}{dx^2} = - \sin x \leq 0$ for $x \in [0, \frac{\pi}{2}]$, $\sin$ is concave over $[0, \frac{\pi}{2}]$. Without loss of generality, let $A \geq B \geq C$. It is clear that $(\frac{A+B+C}{2}, \frac{A+B+C}{2}, 0)$ majorizes $(A, B, C)$, so by Karamata's inequality $\sin A + \sin B + \sin C > \sin \frac{\pi}{2} + \sin \frac{\pi}{2} + \sin 0 = 2$. Lemma 2: Let $S$ be an ellipse with foci $C_1$ and $C_2$ and major axis with endpoints $A_1$, $A_2$, and let $P$ be any point on the major axis of $S$. Then for any point $X$ on the ellipse, $PX \leq \max(PA_1, PA_2)$. Proof: Without loss of generality, let $C_1 = (-c, 0)$, $C_2 = (c, 0)$, $A_1 = (a, 0)$, $A_2 = (-a, 0)$, with $a,c > 0$, $P = (p, 0)$, and let $b = \sqrt{a^2 - c^2}$. It is sufficient to show that the distance from $P$ to $X = (x, b \sqrt{1 - \frac{x^2}{a^2}})$, where $-1 \leq x \leq 1$, is maximized when $x = 1$ or when $x = -1$. For $x \in [-1, 1]$, let $f(x) = PX^2 = (x - p)^2 + b^2 \left(1 - \frac{x^2}{a^2}\right)$. $f(x)$ is a quadratic function with leading coefficient $1 - \frac{b^2}{a^2} > 0$, so $f$ is maximized when $x$ is at an endpoint of $f$'s domain, i.e., when $x = 1$ or $x = -1$. Lemma 3: If $P$ is on the perimeter of $\triangle ABC$, then $f(P) < AB + BC + CA$. Proof: Without loss of generality, let $P$ be on side $BC$. Let $E$ be the ellipse with foci $BC$ passing through $A$. Let $E$ intersect line $BC$ at $A_1$ and $A_2$, with $B$ between $A_1$ and $C$ and $C$ between $B$ and $A_2$. By lemma 2, $PA < \max(PA_1, PA_2)$. We may suppose without loss of generality that $PA < PA_2$. Suppose for the sake of contradiction that $f(P) \geq AB + BC + CA$. Then $f_2(P) = 2PB + PC + AP = BC + PB + AP \geq AB + BC + CA$, whence $PB + AP \geq AB + BC$. Because $AP < AP' = PC + CA_2 = PC + \frac{AB + AC - BC}{2}$, we find that $PB + PC + \frac{AB + AC - BC}{2} > AB + BC$, so $AB + AC + BC > 2AB + 2BC$, so $BC > AB + AC$, which contradicts the triangle inequality. Let $O$ be the circumcenter of $\triangle ABC$. If $M = O$, then $O$ must be in the interior of $\triangle ABC$, whence $\triangle ABC$ must be acute. By lemma 1, $f(O) < AB + BC + CA$, so we may suppose that $M \neq O$. Without loss of generality, let $O$ be the origin. Let $u$ be the vector from $O$ to $M$, and for any nonnegative real $t$ let $\alpha(t) = tu$. (The image of $\alpha$ would be the ray $\overrightarrow{OM}$.) Let $g_i(t) = f_i(\alpha(t))$ for $i = 1,2,3$ and $g(t) = \min(g_1(t), g_2(t), g_3(t))$. We may suppose without loss of generality that $g_1(1) = g(1)$. We claim that $g_1(t) = g(t)$ for all nonnegative $t$. If $g_1(t) \neq g(t)$ for some $t$, then $g_j(t) < g_1(t)$ for some $j$, which we may suppose without loss of generality equals 2. Because $g_2(1) \geq g_1(1)$, the intermediate value theorem implies the existence of some $c \in [\min(t, 1), \max(t, 1)]$ such that $g_1(c) = g_2(c)$. Noting that $t \neq 0$ (since $g_1(0) = g_2(0)$ while $g_1(t) \neq g_2(t)$), we find that $c \geq \min(t, 1) > 0$. Letting $X = \alpha(c)$, we find that $2AX + BX + CX = f_1(X) = f_2(X) = AX + 2BX + CX$, so $AX = BX$. It follows that $X$ must be equidistant from $A$ and $B$, so $X$ must lie on the perpendicular bisector of $AB$. Because $X$ also lies on the line through $\alpha(t)$ and $O$, we find that either $X = O$ or $\alpha(t)$ lies on the perpendicular bisector of $AB$. The former case is impossible since $c \neq 0$. In the latter case, we find that every point on the image of $\alpha$ is equidistant from $A$ and $B$, so $g_1(t) = g_2(t) < g_1(t)$, which is impossible. If $O$ lies inside or on the perimeter of $\triangle ABC$, let ray $\overrightarrow{OM}$ intersect a side of the triangle at $P$, and let $P = \alpha(p)$. Because $g = g_1$ is a convex function over $[0,p]$, the maximum value of $g$ is attained at either $0$ or $p$. Since $M$ is between $O$ and $P$, $0 \leq 1 \leq p$, so $f(M) = g(1) \leq \max(g(0), g(p)) = \max(f(O), f(P))$. By lemmas 1 and 3, $f(O), f(P) < AB + BC + CA$, so our proof in this case is complete. If $O$ lies outside of $\triangle ABC$, let $\overrightarrow{OM}$ intersect two sides of the triangle at $Q$ and $R$. Let $Q = \alpha(q)$ and $R = \alpha(r)$, and suppose without loss of generality that $q < r$. Because $g = g_1$ is a convex function over $[q,r]$, the maximum value of $g$ is attained at either $q$ or $r$. Since $M$ is between $Q$ and $R$, $q \leq 1 \leq r$, so $f(M) = g(1) \leq \max(g(q), g(r)) = \max(f(Q), f(R))$. By lemma 3, $f(Q), f(R) < AB + BC + CA$, so our proof is complete.

Let $D$ be the midpoint of $AB$. Then $M$ lies within either $\triangle CDA$ or $\triangle BDA$. If $M$ lies inside $\triangle BDA$, then $AM +BM < AD+BD < (AB+BC)/2 + BD = s$, where $s$ is the semi-perimeter of $\triangle ABC$. Similarly $\min{(AM+MC,BM+MC)}<s$. The desired result follows.

I finally found a solution after several hours; this problem was much harder than it looked at first WLOG let $AX=\text{min}\{AX, BX, CX\}$ and WLOG $AB\le AC$. Let $\Gamma$ be the circle with center $A$ passing through $X$, and let $\mathcal{E}$ be the ellipse with foci $B,C$ passing through $X$. Let $D=\Gamma \cap AB$ and $E=\Gamma \cap AC$, and finally let $B'\in AC$ s.t $AB=AB'$. We see that $$BD+CD = B'E+CD < B'C+B'D+B'E = B'D+CE=BE+CE$$so if $E$ is in the interior of $\mathcal{E}$, then $D$ is also. However, since $X\in \mathcal{E}$, $D, E$ cannot both be in the interior of $\mathcal{E}$, so $E$ must be outside of $\mathcal{E}$. Thus, $$2AX+BX+CX < 2AX+BE+CE=2AE+BE+CE=AE+AC+BE=AD+AC+B'D < AD+AC+CD < AB+AC+BC$$done. (All inequalities in this solution are applications of triangle inequality, or properties of ellipses.) @Below you are right. The minimality of AX should prevent this though, let me try to fix my solution.

bobthesmartypants wrote: $B'C+B'D+B'E = B'D+CE$ Wrong here: You have to assume that $B'$ is between $C$ and $E$, which may not true.

orl wrote: Let ABC be a triangle and $M$ be an interior point. Prove that \[ \min\{MA,MB,MC\}+MA+MB+MC<AB+AC+BC.\] Let $M$ be a point inside triangle $ABC$. Prove that $$\frac{a+b+c}{2} < MA + MB + MC < a+b+c.$$Let $ABC$ be a triangle and $M$ be a point in $\triangle ABC.$ Denote $a=BC,b=CA,c=AB,$ show that $$a\cdot MA+b\cdot MB+c\cdot MC\le 2\max\{bc,ca,ab\}.$$here

Let's use barycentric coordinates with $A = (1:0:0), B = (0:1:0), C = (0:0:1)$ and $P = (x:y:z).$ Notice that $x, y, z > 0$, and WLOG let $x$ be the largest of $x, y, z.$ Also assume WLOG that $AB \le AC.$ We claim that $2 \cdot PA + PB + PC < AB + BC + CA$, which would suffice. Let $P_1, P_2$ lie on $AB, AC$ respectively so that $P_1P_2 || BC.$ Let $Q_1, Q_2$ be points so that $BP_1PQ_1, CP_2PQ_2$ are parallelograms. Observe that $BP_1 + P_1P > BP, CP_2 + P_2P > CP$ by the Triangle Inequality. These give that $BP_1 + CP_2 + BQ_1 + CQ_2 > BP + CP$, and so we just have to show that $AP_1 + AP_2 + x \cdot BC \ge 2 \cdot PA.$ Claim. $P$ is closer to $P_2$ than to $P_1$. Proof. If not, then we have that $AP \le \max (AP_1, AM)$, where $M$ is the midpoint of $P_1P_2.$ It's easy to see that $AP_2 + AP_1 \ge 2 AP_1$ and we get from Triangle Inequality that $AP_1 + AP_2 \ge 2 AM.$ $\blacksquare$ If $x \ge \frac12$, then $AP_1 + AP_2 + Q_1Q_2 \ge AP_1 + AP_2 + P_1P_2 = (AP_1+P_1P) + (AP_2+P_2P) > 2 \cdot AP.$ Otherwise, suppose that $x < \frac12$, and let $Q$ be the point on segment $P_1P_2$ so that $P_1QQ_2Q_1.$ Lemma. $P$ is on segment $QP_2$, and is closer to $P_2$ than $Q$. Proof. If $P$ is not on segment $QP_2$, then we have from the claim that $AM < AP < AQ < AP_2$, and so $Q_1Q_2 + AP_1 + AP_2 = (AP_1+P_1Q) + AP_2 > AQ + AP_2 > 2 \cdot AP.$ Now, if $P$ is on segment $QP_2$ but is closer to $Q$, letting $N$ be the midpoint of $QP_2$, we have that $Q_1Q_2 + AP_1 + AP_2 = (AP_1+P_1Q) + AP_2 > AQ + AP_2 > 2 \cdot AN \ge 2 \cdot AP$. Hence we must have that the lemma is true. $\blacksquare$ As a result of the lemma, we know that $\frac{P_2Q}{P_1P_2} \le \frac{\frac12 - x}{1-x}$. Since $x < \frac12$, we have that $\frac{\frac12 - x}{1-x} < 1- 2x$, which implies that $\frac{[\triangle AQP_2]}{[\triangle AP_1P_2]} < 1-2x$. However, this means that $y = \frac{[\triangle AQC]}{[\triangle ABC]} < 1 - 2x$. This is absurd, because $z = 1 - x - y \ge 1 - x - x = 1-2x$. $\square$

Let $AM$ intersect $BC$ at $N.$ $AM+BM<AN+BN.$ Suppose WLOG that $N$ is closer to $B$ than $C$ then if $D$ is midpoint of $BC$ then $AN+BN<AD+BD.$ By doubling the median $AD$ and applying triangle inequality we get $2AD<AB+AC$ so we get $AM+BM<s.$ Now, extend $CM$ to intersect $AB$ and apply the same lemma. We get $AM+CM$ or $BM+CM<s.$ This implies result.

Without loss of generality, let $AM \leq BM \leq CM$. Construct $\triangle DEF$, the medial triangle of $\triangle ABC$ as shown below: [asy][asy] size(7cm); pair A = dir(110), B = dir(210), C = dir(330); draw(A--B--C--cycle); pair D = (A+B)/2, E = (B+C)/2, F = (A+C)/2; draw(D--E--F--cycle); label("$A$", A, A); label("$B$", B, B); label("$C$", C, C); label("$D$", D, dir(135)); label("$E$", E, dir(-90)); label("$F$", F, dir(45)); pair M = (0,0); dot(M); draw(A--M--B); draw(C--M); label("$M$", M, dir(30)*1.5); [/asy][/asy] If $M$ lies in trapezoid $AFEB$, we have $$AM + MB < AF + FE + EB = \frac{AB + BC + CA}{2}. (1) $$ If $M$ lies in trapezoid $BDFC$, we have $$BM + MC < BD + DF + FC = \frac{AB + BC + CA}{2}. (2) $$ If $M$ lies in trapezoid $ADEC$, we have $$AM + MC < AD + DE + EC = \frac{AB + BC + CA}{2}. (3)$$ Since $M$ lies in at least 2 of these regions, it is guaranteed that at least one of (2) or (3) is true. If (2) is true, then $$2AM + BM + CM \leq 2BM + 2CM < AB+ BC + CA.$$If (3) is true, then $$2AM + BM + CM \leq 2AM + 2CM < AB+BC+CA.$$Either way, we get that $\text{min}\{AM,BM,CM\} + AM + BM + CM < AB + BC + CA$.

Construction: Draw a equalateral triangle such that $M,$ is the circumcenter. Proof: Let the side length be $2.$ (since all equalateral triangles are congruent we can assume side length) Then, we have $\min\{MA,MB,MC\}+MA+MB+MC=\frac{8}{\sqrt{3}},$ and $AB+AC+BC=6.$ Now, note that $\frac{8}{\sqrt{3}}<6,$ and we are done. note that i am assuming this gives the minimum value of $AB+AC+BC-(\min\{MA,MB,MC\}+MA+MB+MC).$

Add in the medial triangle $DEF$. Of course suppose that WLOG $AM$ is the minimum. It follows that $M$ lies on the parallelogram $AFED$. Also remark that $AEDB$ and $AFDC$ are convex, so thus $AM + BM < \max(AD + BD, BE + AE)$. Likewise, $AM + CM < \max(AD + CD, CF + AF)$. Let's verify these four cases manually. $AD + BD + AD + CD < AB + BC + CA$ iff $2AD < AB + AC$, which is a well known inequality. $AD + BD + CF + AF < AB + BC + CA$ iff $AD + CF < AC + BC/2 + BA/2$, where we are given that $AD + BD > BE + AE$ and $AD + CD < CF + AF$. From the latter we wish to demonstrate \[ 2CF + AF - CD < AC + BC/2 + BA/2 \]which rearranges to $2CF < AC + BC$. This is true. Similarly the other 'mixed' case can be handled in the exact same way. Lastly, $BE + AE + CF + AF < AB + BC + CA$ iff $BE + CF < BC + AB/2 + AC/2$. However this is due to the two inequalities $2BE < AB + BC$ and $2CF < AC + BC$. Done.

Claim: In a trapezoid $ABCD$ with $AB \parallel CD$, Let $M$ be an interior point. Then $MC + MD < AD + AB + BC$. Proof. Let $M'$ be the projection of $M$ onto $AB$. Since $\angle M'MB$ and $\angle M'MC$ are clearly obtuse, from law of cosines $M'D > MD$ and $M'C > MC$. Then \begin{align*} AD + AB + BC &= AD + AM' + M'B + BC \\ &> M'D + M'C \\ &> MD + MC. \ \ \ \ \ \square \end{align*} Let $\triangle DEF$ be the medial triangle of $ABC$. Then $M$ must lie in at least two of the trapezoids $ABDE$, $BCEF$, and $CAFD$. WLOG, suppose $M$ lies in $ABDE$ and $BCEF$. Then \begin{align*} MA + MB &\leq AE + ED + BC = \frac{AB + BC + CA}{2}\\ MB + MC &\leq BF + FE + EC = \frac{AB + BC + CA}{2}. \end{align*}Since $MB \geq \min\{MA, MB, MC\}$, we get \[ \min\{MA,MB,MC\}+MA+MB+MC<AB+AC+BC, \]as desired.

what i actually found the clean sol. Consider the medial triangle $DEF$, we show that if a point $P$ lies in $BCEF$, we can conclude $BP + CP$ is less $\frac a2 + \frac b2 + \frac c2$, and cyclic variants. Proof: We only need to consider the cases where $P$ lies on $EF$ by heightmaxxing, and then by considering varying the minor axis of an ellipse with foci $B,C$, we see that along $EF$ we only need to focus on $E,F$ themselves. This reduces to showing the median is less than the average of the sides it lies adjacent to, we resort to median formula as $m^2 = \frac{2a^2 + 2c^2 - b^2}{4}$, we want $2a^2 + 2c^2 - b^2 < a^2 + c^2 + 2ac$, or $b > |a - c|$, which is true by triangle inequality. Finish: If $M$ lies in $AEF$ or a cyclic variant, we know $\min \{MA, MB, MC\} = MA$ by perpendicular bisectors, so we sum $(MA + MB) + (MA + MC) < a + b + c$. Otherwise, it lies in the medial triangle, and we can sum $2(MA + MB) + 2(MA + MC) + 2(MB + MC) < 3(a + b + c)$, so one of $2MA + MB + MC < a + b + c$ and cyclic variants must be true, as desired.