MHB Proving $PA+PB+PC\geq MA+MB+MC$ in $\triangle ABC$

[Albert1]

M is an inner point of acute $\triangle ABC$

$\angle AMB=\angle BMC=\angle CMA=120^ o$

point P is another point in $\triangle ABC$

Prove :$PA+PB+PC\geq MA+MB+MC$

 

[caffeinemachine]

M is an inner point of acute $\triangle ABC$

$\angle AMB=\angle BMC=\angle CMA=120^ o$

point P is another point in $\triangle ABC$

Prove :$PA+PB+PC\geq MA+MB+MC$

Consider a slightly different question.

Spoiler

Fix a number $l$.
Let $L$ be the locus of all the points $Q$ such that $|QB|+|QC|=l$.
Then $L$ is an ellipse.
Suppose we want to find a point $Q^*$ on $L$ such that $|AQ^*|+|Q^*B|+|Q^*C|$ is minimum.
Imagine a circle whose center is $A$ and whose radius expands with time. At time $t=0$ assume the radius of the circle is $0$.
At some point in time, say $t=t^*$, the circle first comes in contact with the ellipse $L$. Say the radius of this circle is $r^*$ and denote this circle as $C^*$.
One can show that $C^*\cap L$ is a singleton.
Say $C^*\cap L=\{Q^*\}$.
By this construction, we can also see that $Q^*$ is the point on $L$ such that $|AQ^*|+|Q^*B|+|Q^*C|$ is smallest.
By the properties of ellipse, we can also see that $AQ^*$ bisects angle $\angle BQ^*C$.

Now to our problem. Fermat point can be shown to exist in any acute angled triangle. If $F$ is the Fermat point, then $AF$ bisects angle $\angle BFC$, $BF$ bisects angle $\angle AFC$ and $CF$ bisects angle $\angle AFB$. By the above discussion, the inequality in the original question is easily established.

 

[Albert1]

M is an inner point of acute $\triangle ABC$

$\angle AMB=\angle BMC=\angle CMA=120^ o$

point P is another point in $\triangle ABC$

Prove :$PA+PB+PC\geq MA+MB+MC$

Spoiler

View attachment 1851