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.
 A: 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 $\sqrt{2} > \ln 4$.
A: Equivalently, since the exponential function is increasing, we have to prove $\exp(\sqrt2)>4$.
Now,
$$\exp(\sqrt2)=\sum_{n=0}^\infty \frac{(\sqrt2)^n}{n!}>\sum_{n=0}^4 \frac{(\sqrt2)^n}{n!}\\=1+\sqrt2+\frac{2}{2!}+\frac{2\sqrt2}{3!}+\frac{4}{4!}\\=\frac{13+8\sqrt2}{6}$$
Let $x=\dfrac{13+8\sqrt2}{6}$, then $\frac{13}{6}<4$, so both $x-\frac{13}{6}$ and $4-\frac{13}{6}$ are positive, and we can compare their square:
$$(x-\frac{13}{6})^2=\frac{128}{36}$$
While
$$(4-\frac{13}{6})^2=\frac{121}{36}$$
Therefore, $x>4$, so $\exp(\sqrt2)>4$, hence $\sqrt2>\ln 4$.
A: 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.
A: If you are familiar with the trapezoid rule for integration, you can use it to approximate
$\ln 4 =\int_1^4(dx/x)$
Since $f(x) =1/x$ has a positive second derivative everywhere, the piece wise linear function you use to approximate it will lie above the curve, so the numerical integral value you achieve will be an upper bound.
Calculating with six equal intervals, each having width $1/2$, is convenient because then the function values you put into the trapezoid-rule formula will have the form $2/(1+k)$ for whole numbers $k$, which tends to keep denominators relatively small. You eventually get $787/560$ as your upper bound for $\ln 4$. Then $787^2=619369$ versus $2×560^2=627200$, so $(787/560)^2<2$ and the claim is proved.
If you render $\ln 4=2\ln 2$, you can make the calculations simpler. In this case the integral
$\ln 2 =\int_1^2(dx/x)$
would be approximated with the trapezoid rule using just three intervals, which gives an upper bound of $7/10$ (again the second derivative is positive, so the trapezoid rule gives an upper bound). In this method we then have $\ln 4<7/5$ and $(7/5)^2=49/25<2$.
A: Use the inequality $$\log(x) < \sqrt{x} - \frac1{\sqrt x}$$ for $x>1$ and take $x=2$.  This inequality follows from the Cauchy-Schwarz inequality $$\int_1^x \frac{\mathrm ds}s < \left(\int_1^x \frac{\mathrm ds}{s^2}\right)^{\frac 12} \left(\int_1^x \mathrm ds \right)^{\frac 12} $$ or from the trapezium rule for $$\int_1^{\sqrt x} \frac{\mathrm ds}s.$$
A: \begin{align*}
& \int_1^2 (x^{-1/4}-x^{-3/4})^2dx \\\
& =\int_1^2(x^{-1/2} -2x^{-1}+x^{-3/2})dx \\
&=[2x^{1/2}-2\ln(x)-2x^{-1/2}]|_{x=1}^2 \\
&=2(\sqrt2-1)-2\ln(2)-2(\sqrt2/2-1) \\
&=\sqrt{2}-\ln(4) 
\end{align*}
