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AM is a median of a triangle ABC. Is $AB + BC + CA > 2AM$. (Consider the sides of the triangle $\Delta ABM$ and $\Delta AMC$)

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Hint: Here, we will use the property of triangle i.e.., the sum of lengths of any two sides in a triangle should be greater than the length of the third side to prove the given condition $AB + BC + CA > 2AM$.

Complete step-by-step answer:
Given,
AM is a median of a triangle ABC and the triangle is divided into$\Delta ABM{\text{ }}$and $\Delta AMC$.

As we know the property of a triangle i.e.., the sum of lengths of any two sides in a triangle should be greater than the length of the third side. Therefore, let us consider the triangle ABM, we get
$AB + BM > AM \to (i)$
Similarly, from$\Delta AMC$, we get
$AC + MC > AM \to (ii)$
Let us add equation (i) and (ii), we get
  $\Rightarrow AB + BM + AC + MC > AM + AM \\$
  $\Rightarrow AB + AC + (BM + MC) > 2AM \to (iii) \\ $
From the $\Delta ABC$, we know that
$\Rightarrow$ $BM + MC = BC \to (iv)$
So, let us substitute the equation (iv) in equation (iii), we get
$\Rightarrow$ $AB + BC + CA > 2AM$
Hence, equation (i) is proved.

Note: A median of a triangle is a line segment that joins a vertex to the midpoint of the side that is opposite to the vertex bisecting into two equal parts. Here, $BM = MC$ and the area of the triangles ABM and AMC are equal.