Proof and Geometric intuition of $u \leq 2\ln(1+u)$ for $u \in [0,1]$.

Question

How can I prove the inequality $$u \leq 2\ln(1+u)$$ for $u \in [0,1]$.

What is the geometric intuition behind this inequality?

Answer 1

When $u=0,$ $2\ln(1+u)=u=0.$

When $0\le u\le1, \dfrac d {du}(2\ln(1+u)-u)=\dfrac2{1+u}-1=\dfrac{1-u}{1+u}.$

Then $0\le1-u\le1\le1+u,$ so $ \dfrac d {du}(2\ln(1+u)-u)\ge0.$

I.e., $2\ln(1+u)-u$ is $0$ when $u=0$ and non-decreasing when $0\le u\le1;$

therefore, $2\ln(1+u)\ge u$ when $0\le u\le1$.

Answer 2

If you know anything about growth rates and the shapes of these functions you can see it using the following. First draw the shape of the functions on both sides of the inequality - the multiple of 2 and addition of 1 in the log function are irrelevant to the shape of the function. You will see that they must cross at some point where u is positive and after that point the reverse inequality must hold true. Now test some values for the function $2ln(1+u) - u$. If you know that ln2=0.693, you can work it out roughly and see that it is positive and has not yet reached the point where the linear function is above the logarithmic one. If you don't know that, try $u=e-1$ and you will see that the result is still positive and we know that $e-1 > 1$. If you wanted to do a second test, try $u = e^3-1$ and you will see that the result is negative and so the point of intersection must be somewhere between the two values. You can estimate them roughly without any explicit calculations.

Answer 3

A possible geometric way of "seeing" the inequality is interpreting the given expressions as areas under graphs:

$$\ln(1+u) = \int_0^u \frac{dx}{1+x} \stackrel{0\leq u\leq 1}{\geq}\int_0^u \frac{dx}{2} = \frac{u}{2}$$