Find the value of a? Let $a$ and $b$ be  two positive  numbers such that $a\gt b$. Let $G$ be the geometric mean of $a$ and $b$ (that is, $G=\sqrt{ab}$), and $H$ be the Harmonic mean of $a$ and $b$, that is, $$H = \frac{2}{\frac{1}{a}+\frac{1}{b}} = \frac{2ab}{a+b}.$$
If $4G = 5H$, what is the value of $a$?

According to the given information we have,
$$4\sqrt{ab} = 5(\frac{2ab}{a+b})$$
$$(a+b) =\frac{5}{2}\sqrt{ab})$$
$$\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}} = \frac{5}{2}$$
Let , $t$ $ =$ $\sqrt{\frac{a}{b}}$
$$[t+\frac{1}{t}=\frac{5}{2}].....................Eq(1)$$
A clever person will immediately infer that $t=\frac{1}{2}$
But if its a subjective question we have to justify that also, so
$$({t+\frac{1}{t}})^2=\frac{25}{4}$$
$$t^2 +\frac{1}{t^2} = \frac{17}{4}$$
NOW,
$$(t-\frac{1}{t})^2= t^2+\frac{1}{t^2} -2 =\frac{9}{4}$$
$$[t-\frac{1}{t}=\frac{3}{2}   ]   .....................Eq(2)$$
neglecting the negative value as we know that$ L.H.S.>0$ , since $,t>0$
From Eq(1) and Eq(2) we have $t=4$, hence $a=4b$
