If \(yz=a^2\) prove that \[ \frac{1}{a+y} + \frac{1}{a+z} = \frac{1}{a}.\]

Solving \(yz=a^2\) for \(y\), we obtain

\[ y=\dfrac{a^2}{z},\]

which we can substitute into the left-hand side of the given expression. This leads us to

\[\begin{align*} \frac{1}{a+y} + \frac{1}{a+z} &= \frac{1}{a+\dfrac{a^2}{z}} + \frac{1}{a+z}\\ &=\dfrac{z}{az+a^2} + \dfrac{a}{a(a+z)}\\ &=\dfrac{a+z}{a(a+z)}=\dfrac{1}{a}, \end{align*}\]

as required.

By using this result, or otherwise, find the numerical values of

  1. \(\dfrac{1}{1+x^k} + \dfrac{1}{1+x^{-k}}\);

Comparing this expression with the one above leads to setting \(a=1\), \(x^k=y\) and \(x^{-k}=z\). These values clearly fulfil \(yz=a^2\), and hence give us

\[\frac{1}{1+x^k} + \frac{1}{1+x^{-k}} = 1.\]

  1. \(\dfrac{1}{7+\sqrt{62}-\sqrt{13}} + \dfrac{1}{7+\sqrt{62}+\sqrt{13}}\);

We have \((\sqrt{62}-\sqrt{13})(\sqrt{62}+\sqrt{13})=62-13=49=7^2\), which means we can we take \(a=7\), \(y=\sqrt{62}-\sqrt{13}\) and \(z=\sqrt{62}+\sqrt{13}\).

Thus \[\frac{1}{7+\sqrt{62}-\sqrt{13}} + \frac{1}{7+\sqrt{62}+\sqrt{13}}=\frac{1}{7}.\]

  1. \(\dfrac{2}{5(2+5\log_b c)} + \dfrac{1}{5+2\log_c b}\);

We can write this as

\[\frac{1}{5+\dfrac{25}{2}\log_b c} + \frac{1}{5+2\log_c b}.\]

This looks rather like the expression at the start of the question, so let us try \(a=5\), \(y=\dfrac{25}{2}\log_b c\) and \(z=2\log_c b\). Then \[yz=\dfrac{25}{2}\log_b c\times2\log_c b = 25\log_b c\times\log_c b = 25 = a^2 .\]

We have used the law of logarithms for changing bases which states that \[\log_a b=\log_a c\times \log_c b .\]

So we can state that \[\frac{2}{5(2+5\log_b c)} + \frac{1}{5+2\log_c b}=\frac{1}{5}.\]