Solving given log inequality for another bound of x

Question

Here's a log inequality I found from Sedgewicks Algorithms in C : $$ (lnx)^2 < 2\sqrt x $$ Its obviously need to be solved using Lambert W function.. So I proceeded as :
Taking natural logarithm $$ 2 ln(lnx) < ln2 + \frac 1 2 lnx $$ Allowed since its monotonically increasing, Lets suppose : $$ lnx = m $$ for simplicity.And dividing by 2 $$ ln(m) < \frac {ln2} 2 + \frac m 4$$ Taking exponentiation, $$ m < \sqrt 2. e^{m/4} $$ $$ m e^ {\frac {-m} 4} < \sqrt 2 $$ $$ \frac {-m} 4 e^ {\frac {-m} 4} > \frac {-\sqrt 2} 4 $$ Using Lambert W Function (+ve branch) on both sides, $$ \frac {-m} 4 > W(\frac {-1} {2\sqrt 2}) $$ $$ m < -4 W(\frac {-1} {2\sqrt 2})$$ And,
$$ x < e^{-4W( \frac {-1} {2\sqrt 2} ) } $$ Similarly, taking -ve branch we will get, $$ x > e^{-4 W_{-1}(\frac {-1} {2\sqrt 2})}$$ Pluging the approximate value we get, from above $$ x < 19.6077 $$$$ x > 187.867 $$ Its not complete solution and I'm still missing lower bound of x for which : $$ y < x < 19.6077 $$ How can y be found ? Is there any way to extract it from above process?

Answer 1

There are three positive real solutions of $\ln(x)^2 = 2 \sqrt{x}$: in increasing numerical order they are

$$ 64 W(\sqrt{2}/4)^4, 64 W(-\sqrt{2}/4)^4, 64 W_{-1}(-\sqrt{2}/4)^4 $$ where $W$ is the principal branch of Lambert W and $W_{-1}$ the "$-1$" branch.

Your inequality is true for $64 W(\sqrt{2}/4)^4 < x < 64 W(-\sqrt{2}/4)^4$ and $x > 64 W_{-1}(-\sqrt{2}/4)^4$