A solution was given in http://www.artofproblemsolving.com/Forum/viewtopic.php?t=273 post #6 by Johann Peter Dirichlet: "Let ABC be a triangle and X an interior point of ABC. Show that at least one of the angles XAB, XBC, XCA is less than or equal to 30 degrees." The solution is obvious with the Brocard point! The anterior messages contains all necessary theorems. P is the Brocard poin of the triangle ABC. If X=P then aren't nothing to prove. If X<>P then X is in the interior of one of the triangles PAB,PBC,PCA (wlog PAB).Then the problem is obvious (connect the point X to point A and verify the angle).

Here's a solution that I found right now. It is very late at my place, so I will skip the obvious parts. Solution. Assume to the contrary that all the angles are greater than $ 30^{\circ}$. Let $ PA=x,PB=y,PC=z$. Then it follows that $ x^2=z^2+b^2-2bz\cos\angle PCA>z^2+b^2-bz\sqrt{3}$. Similarly, we find anologous inequalities. Adding them, we obtain $ a^2+b^2+c^2<\sqrt{3}(bz+ay+cx)$. But $ [ABC]=\frac{1}{2}\sum cx\sin\angle PAB>\frac{ay+bz+cx}{4}$. Thus, $ a^2+b^2+c^2<4\sqrt{3}[ABC]$. This is a contradiction to a well-known inequality (an old IMO problem that can be easily proven using Heron's formula along with AM-GM). $ \Box$

Let $ d(P,AB)=x$ , $ d(P,BC)=y$ , $ d(P,CA)=z$ lemma. $ PA + PB + PC \geq 2(x + y + z)$ At least one of $ PA\geq 2x$ , $ PB\geq 2y$ , $ PC\geq 2z$ holds by lemma. if we assume $ PA\geq 2x$ , $ \frac {x}{PA}\leq\frac {1}{2}$ and $ \angle{PAB}\leq 30^{\circ}$

erdos-mordell inequality (when converted to trig form0 gives an immediate contradiction when we assume that all marked angles are at least 30.

Here's a very simple solution that I found for this problem: If we draw a Triangle and interior point P, we label the vertices A, B, C anyway we want WLOG. Since the labeling of the vertices of the triangle can be interchanged, it suffices to show that at least one of the 6 angles $ \,\angle PAB,\;\angle PBC,\;\angle PCA,\;\angle PBA ,\;\angle PCB ,\;\angle PAC\,$ is less than or equal to 30. We prove this statement by contradiction. First, we assume that all six of these angles are greater than 30. Then, the sum of these angles must be greater than 180, which is a contradiction since the sum of these six angles must be 180. Thus, at least one of these 6 angles must be less than or equal to 30. We can label the vertices A, B, and C so that either $ \,\angle PAB,\;\angle PBC,\;\angle PCA\,$ is one of these six angles. Can someone look through and see if I made any mistakes in this proof?

Assume that $\,\angle MAB, \,\angle MBC, \,\angle MCA$ are all greater than $30^{\circ}$. By $\textbf{sine Ceva},$ $\sin(MAC)\sin(MBA)\sin(MCB)$ = $\sin(MAB)\sin(MBC)\sin(MCA) > sin^3(30^{\circ})=\frac{1}{8}$. Also, from other side, $\,\angle MAC + \,\angle MBA + \,\angle MCB < 180^{\circ} -3*30^{\circ} = 90^{\circ}$. Note that the function $f(x)=\log(\sin(x))$ is concave on $[0,\pi]$. Applying Jensen, we get a contradiction here.

Trivialized by Jensen's But more trivialized by Erdos-Mordell inequality: Let the projections of $M$ onto sides $BC,AC,$ and $AB$ be $X,Y,$ and $S$, respectively. By Erdos-Mordell, $$MA+MB+MC \geq 2(MX+MY+MS).$$This implies that at least one of $MA \geq 2MS$, $MB \geq 2MX$, or $MC \geq 2MY$ must be true. WLOG, let $MA \geq 2MS$. Since in right $\triangle SAM$ we have $\sin SAM \leq \frac{MS}{MA}=\frac{1}{2}$, which implies $\angle SAM \leq 30^\circ$ and we are done. $\mathbb{S.A.M.!}$

Same as everyone else.

Suppose, for the sake of contradiction, that all angles are greater than $30^\circ$ and less than $150^\circ$. Application of Erdos-Mordell gives that \begin{align*} PA+PB+PC&\ge2(PD+PE+PF)\\ &=2PA\sin\angle PAB+2PB\sin\angle PBC+2PC\sin\angle PCA\\ &>PA+PB+PC, \end{align*}a contradiction; thus, one of $\angle PAB$, $\angle PBC$, and $\angle PCA$ is at most $30^\circ$ or at least than $150^\circ$. However, if one of those angles, without loss of generality let it be $\angle PAB$ is at least $150^\circ$, then $\angle PBC<\angle B\le 30^\circ$, so there exists an angle at most $30^\circ$.