I have this inequality that I deduce from the problem:
$$ \frac{3}{a+b} < \frac{1}{2} \left(\frac{1}{a} + \frac{1}{b}\right) $$
For the sake of problem: assuming $a,b > 0$ and $a \neq b$. How do I solve for the relation? (I was thinking this relates to AM - GM inequality).
We have:
$$0<\frac{1}{a}+\frac{1}{b}-\frac{6}{a+b}=\frac{a^2-4ab+b^2}{ab(a+b)}$$
To hold, we need that $a^2-4ab+b^2>0$. Dividing by $b^2$ we get the following
$$\frac{a^2}{b^2}-4\frac{a}{b}+1>0$$
Let $t = \dfrac{a}{b}$. Substituting we get the following second degree inequality $\quad t^2-4t+1 > 0 \quad$ which is solved when $t < 2-\sqrt{3}$ or $t > 2+\sqrt{3}$.
In conclusion the two possible solutions are: $$a < (2-\sqrt{3})b \\ a>(2+\sqrt{3})b$$