<s>[INST] Without using approximation, prove that $ \ln(4)<\sqrt{2}$. 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.
 [/INST] The function $f(x) = \frac{1}{x}$ is convex. Therefore, we have $\frac{(f(\sqrt{2}) + f(1))(\sqrt{2} - 1)}{2} > \int\limits_1^{\sqrt{2}} \frac{1}{x} dx$ by trying to approximate the region of integration as a trapazoid.
Evaluating both sides gives us $\frac{1}{2}(\frac{\sqrt{2}}{2} + 1)(\sqrt{2} - 1) > \ln \sqrt{2}$. Multiply both sides by 4 to get $(\sqrt{2} + 2)(\sqrt{2} - 1) > \ln 4$.
The left-hand side simplifies to $\sqrt{2}$. So we have $\sqrt{2} > \ln 4$.
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