Two proofs involving Harmonic Mean If H is the harmonic mean between $a$ and $b$,then show that
$$\frac{1}{H-a}+\frac{1}{H-b} = \frac{1}{a} + \frac{1}{b}$$ and $$\frac{H+a}{H-a}+\frac{H+b}{H-b} = 2$$
I substituted  $\displaystyle H = \frac{2ab}{a+b}$, then tried some algebraic manipulation, but I am not getting there. 
Are they even valid? If yes, could somebody give me some ideas how to approach these?

I don't know why you aren't getting this. You have 
\begin{align*}
\frac{1}{H-a} + \frac{1}{H-b} &= \frac{1}{\frac{2ab}{a+b} -a} + \frac{1}{\frac{2ab}{a+b}-b} \\ &= \frac{a+b}{ab-a^{2}} + \frac{a+b}{ab-b^{2}} \\ &= (a+b) \cdot \biggl[ \frac{1}{a(b-a)} + \frac{1}{b(a-b)}\biggr] \\
\end{align*}
I think this should help you out. Try taking the factor $(a-b)$ common. For the next problem also again substitute the value of $H$ and try doing manipulations. Be patient, and be careful, with your calculations, you shall arrive at the result.
