# 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.

• Are you trying to avoid differentiating? Jan 12, 2022 at 19:57
• Seems like it will involve the Lambert function is you try to find explicit roots Jan 12, 2022 at 20:02
• See this post. Or is this approximation? Some answers not, I think. Do the same for $\log(4)$ instead of $\log(3)$. Jan 12, 2022 at 20:04
• Wolfram says $2.7^{1.4} > 4$, so you might be able to prove that. (Not sure if that counts as approximation) Jan 12, 2022 at 20:29
• Some context would help. Is this for a numerical analysis class in which you are allowed to use things like Taylor expansions with remainders, or alternating series remainder result, or approximate integration methods (for upper and lower Riemann sums), etc.? (If so, then your title is not appropriate.) Is this something in which the expectation is that you apply some kind of exotic contest math inequality? Jan 12, 2022 at 20:39

\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*}

• It is a clever approach. It would be improved by sharing, if not the secret of how you came up with this, at least a step or two of the derivation of the result (definite integral). Jan 12, 2022 at 22:41
• Added the missing steps. Jan 13, 2022 at 13:26

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 $$\int_1^{\sqrt x} \frac{\mathrm ds}s.$$

• (+1) I had written an equivalent answer, but substituting $x\mapsto1+x$, which gives $\log(1+x)\le\frac{x}{\sqrt{1+x}}$. This did not immediately look the same, but it is, so I deleted it.
– robjohn
Jan 15, 2022 at 14:47
• (+1) I found $x \log{x} < (x - 1) + \frac{1}{2} (x - 1)^2$ via the Taylor expansion of $x \log{x}$ around $x = 1$. This is the same as your inequality upon expanding and substituting $x \mapsto \sqrt{x}$.
– Ant
Feb 1, 2022 at 21:04

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 $$\sqrt{2} > \ln 4$$.

• An inprovement of my trapezoid rule approach. "I am the primitive of the method I invented." -- Paul Cézanne Jan 13, 2022 at 13:17

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 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.

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 $$(7/5)^2=49/25<2$$.

• you can use it to approximate -- This seems not allowed according to the OP's title, but perhaps the title was not correctly (or sufficiently precisely) stated. Jan 12, 2022 at 20:41
• I am defining a rigorous upper bound. I think that would be allowed. Jan 12, 2022 at 20:47
• I also suspect it would be allowed, but more clarification from the OP would help. When I see a bald question like this, I feel like I'm shooting in the dark regarding what is appropriate -- only school level math? beginning calculus? undergraduate numerical analysis? contest type math in which we're allowed to bring in little known techniques analogous to playing scrabble? advanced numerical techniques? graduate level variations of things like the Minkowski inequality? Jan 12, 2022 at 20:56