If a,b,c are in H.P. then

$$\begin{gathered} \frac{1}{\mathrm{a}},\frac{1}{\mathrm{b}},\frac{1}{\mathrm{c}}\text{ are in A.P. } \\ \Rightarrow \frac{a+b+c}{a},\frac{a+b+c}{b},\frac{a+b+c}{c}\text{ are in A.P. } \\ \Rightarrow \frac{\mathrm{b}+\mathrm{c}}{\mathrm{a}},\frac{\mathrm{c}+\mathrm{a}}{\mathrm{b}},\frac{\mathrm{a}+\mathrm{b}}{\mathrm{c}}\text{ are in A.P. (Subtracting }1\text{ from each term) } \\ \Rightarrow \frac{\mathrm{b}+\mathrm{c}}{\mathrm{a}}-1,\frac{\mathrm{c}+\mathrm{a}}{\mathrm{b}}-1,\frac{\mathrm{a}+\mathrm{b}}{\mathrm{c}}-1\text{ are in A.P. (Subtracting }1\text{ from each term) } \\ \Rightarrow \frac{\mathrm{b}+\mathrm{c}-\mathrm{a}}{\mathrm{a}},\frac{\mathrm{c}+\mathrm{a}-\mathrm{b}}{\mathrm{b}},\frac{\mathrm{a}+\mathrm{b}-\mathrm{c}}{\mathrm{c}}\text{ are in A.P. } \end{gathered}$$

$$\begin{gathered} \mathrm{a},\mathrm{b},\mathrm{c} \mathrm{are} \mathrm{in} \mathrm{H}.\mathrm{P} \\ \Rightarrow \mathrm{b}=\frac{2\mathrm{ac}}{\mathrm{a}+\mathrm{c}} \\ \frac{1}{b-a}+\frac{1}{b-c}=\frac{2b-(a+c)}{(b-a)(b-c)}=\frac{2b-(a+c)}{b^2-b(a+c)+ac} \\ =\frac{2b-2ac/b}{b^2-b\left(2ac\right)b+ac}=\frac{2}{b}\cdot \frac{b^2-ac}{b^2-ac}=\frac{2}{b} \end{gathered}$$

a,b,c are in H.P

$$\begin{gathered} b=\frac{2ac}{a+c}\Rightarrow ab+bc=2ac \\ \Rightarrow 2ab+2bc=4ac \\ \Rightarrow 2ab+2bc+b^2=4ac+b^2 \\ \Rightarrow b^2=4ac-2ab-2bc+b^2 \\ \Rightarrow b^2=2a\left(2c-b\right)-b\left(2c-b\right) \\ \Rightarrow b^2=\left(2c-b\right)\left(2a-b\right) \\ \Rightarrow {\left(\frac{b}{2}\right)}^2=\left(a-\frac{b}{2}\right)\left(c-\frac{b}{2}\right) \end{gathered}$$