Question

a, b, c are positive real numbers which are in HP. Then \[\dfrac{a+b}{2a-b}+\dfrac{c+b}{2c-b}\] is
(a) \[\ge 4\]
(b) \[\le 4\]
(c) equal to 0
(d) none of these

Answer 1

Hint: a, b and c are in H.P. The relation connecting them with, \[b=\dfrac{2ac}{\left( a+c \right)}\]. Now simplify the given expression and substitute 2ac in place of b (a + c) on further simplification put \[\dfrac{2ac}{b}\] on (a + c). Now simplify it using the expression, \[\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}\] and find the expression.

Complete step-by-step answer:
It is given that a, b, c are in Harmonic progression (HP). A HP is defined as a sequence of real number which is determined by taking the reciprocals of the arithmetic progression that does not contain zero.
As a, b, c are in HP, they are can be written as,
\[\Rightarrow b=\dfrac{2ac}{a+c}\] - (1)
We have been asked to find the value of,
\[\Rightarrow \dfrac{a+b}{2a-b}+\dfrac{c+b}{2c-b}\]
Let us simplify the expression,
\[\begin{align}
  & =\dfrac{\left( a+b \right)\left( 2c-b \right)+\left( c+b \right)\left( 2a-b \right)}{\left( 2a-b \right)\left( 2c-b \right)} \\
 & =\dfrac{2ac-ab+2bc-{{b}^{2}}+2ac-bc+2ab-{{b}^{2}}}{4ac-2ab-2bc+{{b}^{2}}} \\
\end{align}\]
\[=\dfrac{4ac+ab+bc-2{{b}^{2}}}{4ac-2ab-2bc+{{b}^{2}}}=\dfrac{4ac+b\left( a+c \right)-2{{b}^{2}}}{4ac-2b\left( a+c \right)+{{b}^{2}}}\] - (2)
From (1), \[b\left( a+c \right)=2ac\].
Hence let us replace \[b\left( a+c \right)\] from (2) with 2ac.
\[\begin{align}
  & \therefore \dfrac{4ac+b\left( a+c \right)-2{{b}^{2}}}{4ac-2b\left( a+c \right)+{{b}^{2}}}=\dfrac{4ac+2ac-2{{b}^{2}}}{4ac-2\times 2ac+{{b}^{2}}} \\
 & =\dfrac{6ac-2{{b}^{2}}}{4ac-4ac+{{b}^{2}}}=\dfrac{6ac-2{{b}^{2}}}{{{b}^{2}}}=\dfrac{6ac-2{{b}^{2}}}{{{b}^{2}}} \\
 & =\dfrac{6ac}{{{b}^{2}}}-2 \\
\end{align}\]
From (1) we can form, \[\dfrac{2ac}{b}=a+c\]. Thus put the same in the above expression,
\[\Rightarrow \dfrac{3}{b}\times \left( \dfrac{2ac}{b} \right)-2=\dfrac{3}{b}\left( a+c \right)-2\]
As a, b and c are in HP we can write, \[\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}\] \[\Rightarrow \dfrac{1}{b}=\dfrac{1}{2}\left( \dfrac{1}{a}+\dfrac{1}{c} \right)\].
\[\therefore \dfrac{3}{b}\left( a+c \right)-2\]
\[\therefore \dfrac{3}{b}\left( a+c \right)-2=3\left( a+c \right)\times \dfrac{1}{b}-2=3\left( a+c \right)\times \dfrac{1}{2}\left( \dfrac{1}{a}+\dfrac{1}{c} \right)-2\]
\[\begin{align}
  & \therefore \dfrac{3}{b}\left( a+c \right)-2=\dfrac{3}{2}\left( a+c \right)\left[ \dfrac{1}{a}+\dfrac{1}{c} \right]-2 \\
 & \therefore \dfrac{3}{b}\left( a+c \right)-2=\dfrac{3}{2}\left[ \dfrac{a}{a}+\dfrac{a}{c}+\dfrac{c}{a}+\dfrac{c}{c} \right]-2 \\
\end{align}\]
\[\begin{align}
  & \therefore \dfrac{3}{b}\left( a+c \right)-2=\dfrac{3}{2}\left[ 1+\dfrac{a}{c}+\dfrac{c}{a}+1 \right]-2=\dfrac{3}{2}\left[ 2+\dfrac{a}{c}+\dfrac{c}{a} \right]-2 \\
 & \therefore \dfrac{3}{b}\left( a+c \right)-2=3+\dfrac{3a}{2c}+\dfrac{3c}{2a}-2 \\
\end{align}\]
\[\therefore \dfrac{3}{b}\left( a+c \right)-2=1+\dfrac{3}{2}\left( \dfrac{a}{c}+\dfrac{c}{a} \right)\] - (2)
Now let us find the arithmetic mean relation on \[\dfrac{a}{c}\] and \[\dfrac{c}{a}\].
\[\begin{align}
  & \Rightarrow \dfrac{AM}{2}\ge \sqrt{GM} \\
 & \Rightarrow \dfrac{\left( \dfrac{a}{c}+\dfrac{c}{a} \right)}{2}\ge \sqrt{\dfrac{a}{c}\times \dfrac{c}{a}}\Rightarrow \dfrac{a}{c}+\dfrac{c}{a}\ge 2 \\
\end{align}\]
i.e. \[1+\dfrac{3}{2}\left( \dfrac{a}{c}+\dfrac{c}{a} \right)\ge 1+\dfrac{3}{2}\times 2=1+3=4\]
\[\therefore \] Then the expression will be greater than or equal to 4.
\[\therefore \] Option (a) is the correct answer.

Note: The basic for this solution are the formula, \[b=\dfrac{2ac}{a+c}\] and \[\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}\]. Be careful while substituting and don’t misplace the variables. After finding the required expression don’t forget to find the arithmetic mean of \[\dfrac{a}{c}\] and \[\dfrac{c}{a}\].