Geometric Substitution: \( a=y+z \), \( b=z+x \), \( c=x+y \)

Problem 1
 
If \( a \), \( b \), and \( c \) are the lengths of the sides of a triangle, prove that \[ \frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{a+b-c}\geq 3.\]

Let us first discuss this problem for a bit. Many problems look like this one and there is a trick for handling them. We have to use the information that \( a \), \( b \), \( c \) are sides of a triangle. Till now, all we know about sides is that each of them is less than the sum of the other two. This gives us three inequalities and it is very hard to utilize them. We will use the following trick. Let \( ABC \) be the triangle and assume that the incircle touches the sides at \( A^{\prime\prime} \), \( B^{\prime\prime} \), \( C^{\prime\prime} \). Then \( AB^{\prime\prime}=AC^{\prime\prime} \) and let us denote this quantity by \( x \). Similarly let \( y=BC^{\prime\prime}=BA^{\prime\prime} \) and \( z=CB^{\prime\prime}=CA^{\prime\prime} \). Then \( a=y+z \), \( b=z+x \), \( c=x+y \). The wonderful thing about this substitution is the following: Positive real number \( a \), \( b \), \( c \) are the sides of a triangle if and only if there are positive real numbers \( x \), \( y \), \( z \) such that \( a=x+y \), \( b=y+z \), \( c=z+x \).

Problem 2
 
Prove that \[ a^2(s-a)+b^2(s-b)+c^2(s-c)\leq \frac32abc.\]


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