Show that $\displaystyle \frac{2+\ln a }{2}\lt\frac{a-1}{\ln a} \lt \frac{1+a}{2}$ becomes $\displaystyle \frac{2(a-1)}{a+1}\lt\ln a \lt -1 + \sqrt{2a-1}$
The closest I can get is $$ \frac{2(a-1)}{a+1}\lt\ln a\lt\frac{2(a-1)}{2+\ln a}.$$ But I did this by flipping both sides and reversing the inequality, I lack the relevant knowledge to know if this was a valid move.
I've often been told off for not giving context to my questions so here it is - a STEP paper question:
The first part i) I can solve and I already have a solution to the second part ii):
Was there any way of reaching that result without having to separate out the inequality? In other words can it be done by performing operations on both sides of $$ \frac{2(a-1)}{a+1}\lt\ln a\lt\frac{2(a-1)}{2+\ln a}.$$
Best regards, thank you.
$\ln a < \frac{2(a-1)}{2+\ln a} <=> (\ln a)^2+2\ln a<2(a-1) <=> (\ln a)^2+2\ln a+1<2(a-1)+1 <=> (\ln a+1)^2<2a-1 => \ln a < \sqrt{2a-1}-1$,
Note that $\ln a$ is positive, for the above multiplication by $\ln a +2$, since $a$ is bigger than $1$.