To show that
[tex]\[
\frac{a-A}{a-H} \times \frac{b-A}{b-H} = \frac{A}{H}
\][/tex]
we start by defining the Arithmetic Mean (A) and the Harmonic Mean (H) of two quantities [tex]\(a\)[/tex] and [tex]\(b\)[/tex].
1. Calculate the Arithmetic Mean (A):
[tex]\[
A = \frac{a + b}{2}
\][/tex]
2. Calculate the Harmonic Mean (H):
[tex]\[
H = \frac{2ab}{a+b}
\][/tex]
Next, we will substitute [tex]\(A\)[/tex] and [tex]\(H\)[/tex] into the left-hand side of the equation we want to show.
3. Substitute [tex]\(A\)[/tex] and [tex]\(H\)[/tex] into the left-hand side:
[tex]\[
\frac{a-A}{a-H} \times \frac{b-A}{b-H}
\][/tex]
First substitute for [tex]\(A\)[/tex]:
[tex]\[
a - A = a - \frac{a + b}{2} = \frac{2a - (a + b)}{2} = \frac{a - b}{2}
\][/tex]
[tex]\[
b - A = b - \frac{a + b}{2} = \frac{2b - (a + b)}{2} = \frac{b - a}{2}
\][/tex]
Then substitute for [tex]\(H\)[/tex]:
[tex]\[
a - H = a - \frac{2ab}{a + b}
\][/tex]
To simplify [tex]\(a - H\)[/tex], we find a common denominator:
[tex]\[
a - H = a - \frac{2ab}{a + b} = \frac{a(a + b) - 2ab}{a + b} = \frac{a^2 + ab - 2ab}{a + b} = \frac{a^2 - ab}{a + b} = \frac{a(a - b)}{a + b}
\][/tex]
Similarly,
[tex]\[
b - H = b - \frac{2ab}{a + b} = \frac{b(a + b) - 2ab}{a + b} = \frac{ab + b^2 - 2ab}{a + b} = \frac{b^2 - ab}{a + b} = \frac{b(b - a)}{a + b}
\][/tex]
Substitute these expressions back into the fractions:
[tex]\[
\frac{a - A}{a - H} = \frac{\frac{a - b}{2}}{\frac{a(a - b)}{a + b}} = \frac{a - b}{2} \times \frac{a + b}{a(a - b)} = \frac{a + b}{2a}
\][/tex]
[tex]\[
\frac{b - A}{b - H} = \frac{\frac{b - a}{2}}{\frac{b(b - a)}{a + b}} = \frac{b - a}{2} \times \frac{a + b}{b(b - a)} = \frac{a + b}{2b}
\][/tex]
Now multiply these fractions together:
[tex]\[
\frac{a - A}{a - H} \times \frac{b - A}{b - H} = \frac{a + b}{2a} \times \frac{a + b}{2b} = \frac{(a + b)^2}{4ab}
\][/tex]
4. Calculate the right-hand side [tex]\( \frac{A}{H} \)[/tex]:
[tex]\[
\frac{A}{H} = \frac{\frac{a + b}{2}}{\frac{2ab}{a + b}} = \frac{a + b}{2} \times \frac{a + b}{2ab} = \frac{(a + b)^2}{4ab}
\][/tex]
5. Compare the left-hand side and the right-hand side:
[tex]\[
\frac{(a + b)^2}{4ab} = \frac{(a + b)^2}{4ab}
\][/tex]
Since the left-hand side equals the right-hand side, we have shown that:
[tex]\[
\frac{a-A}{a-H} \times \frac{b-A}{b-H} = \frac{A}{H}
\][/tex]
