TITLE: Without using approximation, prove that $ \ln(4)<\sqrt{2}$.
QUESTION [11 upvotes]: I have been asked to prove the inequality $$\ln(4)<\sqrt{2}$$
without using the fact that
$$\ln(4)\approx 1.38 \text{ and }\;\sqrt{2}\approx 1.41$$
I defined at $ [1,2] $ the function
$$f(x)=x^3\ln(x)-1$$
and tried to see if $ f(\sqrt{2})<0 $ but i need to know where the sign of $ f(x) $ changes.
Thanks in advance.

REPLY [4 votes]: A known definite integral trick (Using $\int_0^1 \frac{x^m (1 - x)^n}{1 + x}\mathrm{d} x$):
We have
$$0 \le \int_0^1 \frac{x(1 - x)^3}{1 + x}\mathrm{d} x
= \int_0^1 \left(-x^3 + 4x^2 - 7x + 8 - \frac{8}{1 + x}\right)\mathrm{d} x
= \frac{67}{12} - 8\ln 2$$
which results in
$$\frac{67}{96} \ge \ln 2.$$
Thus,
$$\ln 4 < \sqrt2 \iff 2\ln 2 < \sqrt2 \Longleftarrow 2\cdot \frac{67}{96} < \sqrt2 \iff 4\cdot \frac{67^2}{96^2} < 2$$
which is true.