 QUESTION

# If $\dfrac{1}{a\left( b+c \right)},\dfrac{1}{b\left( c+a \right)},\dfrac{1}{c\left( a+b \right)}$ are in H.P. , then a, b, c are also in H.P.

Hint: We need to first convert the given three terms in H.P to the terms in A.P and then simplify the equation using the condition of A.P which gives the relation between a, b, c.Comparing this relation with condition of H.P then we can say a, b, c are also in H.P.

Three terms x, y, z are in H.P then the terms $\dfrac{1}{x},\dfrac{1}{y},\dfrac{1}{z}$ are said to be in A.P
If x, y, z are in A.P then $y=\dfrac{x+z}{2}$
If x, y, z are in H.P then $y=\dfrac{2xz}{x+z}$.
Now, let us consider the terms given in the question.
$\Rightarrow \dfrac{1}{a\left( b+c \right)},\dfrac{1}{b\left( c+a \right)},\dfrac{1}{c\left( a+b \right)}$
As given that these terms are in H.P. we can say that the inverse of these functions are said to be in A.P.
$\Rightarrow a\left( b+c \right),b\left( c+a \right),c\left( a+b \right)$
These terms are said to be in A.P
From the conditions of A.P we have that when three terms x, y, z are in A.P then we can express them as:
$y=\dfrac{x+z}{2}$
This is also called arithmetic mean.
Now, from the above three terms we have we can write them as,
$\Rightarrow b\left( c+a \right)=\dfrac{a\left( b+c \right)+c\left( a+b \right)}{2}$
Now, let us multiply with 2 on both sides.
$\Rightarrow 2b\left( c+a \right)=a\left( b+c \right)+c\left( a+b \right)$
Let us now multiply them inside and expand the equation.
$\Rightarrow 2bc+2ba=ab+ac+ca+cb$
Let us now rearrange the terms and subtract the common terms.
$\Rightarrow 2bc-bc+2ba-ba=ac+ac$
Now, on further simplification we get,
$\Rightarrow bc+ba=2ac$
Now, we can rewrite this as:
$\Rightarrow b\left( a+c \right)=2ac$
Let us now divide with $a+c$ on both sides.
$\Rightarrow b=\dfrac{2ac}{a+c}$
As we already know that from the condition of H.P that when three terms x, y, z are in H.P then we can write them as:
$y=\dfrac{2xz}{x+z}$
This is also called the harmonic mean of the three numbers.
Hence, a, b, c are in H.P

Note: Instead of writing the given terms of H.P in A.P we can directly apply the condition of H.P which is also known as the harmonic mean and then simplify it accordingly. But, it will be a long process and a bit more confusing. Both the methods give the same result.
It is important to note that while solving the equations using appropriate conditions we should not neglect any one of the terms. Because neglecting any of the terms changes the final equation which cannot be shown as either H.P or A.P.