# Inequality - L'Hôpital's rule, logarithms

Let's consider a function: (EDIT here: the number "2" was lacking)

$\mathcal{B}(x) = \frac{2}{x^2} \big( (1+x) log(1+x) - x \big)$

I need to prove inequality:

$(1 + \frac{1}{3}x)\, \mathcal{B}(x) \geq 1 \, , \qquad x>0 \, ,$

using a hint: "Apply L'Hôpital's rule twice to reduce left-hand side to $\frac{1 + \frac{1}{3}x ^\ast}{1 + x^\ast} + \frac{2}{3} log (1 + x^\ast)$ for some $x^\ast$ less that $x$. Then use: $\quad log(1+x^\ast) \geq \frac{x^\ast}{1 + x^\ast} \,$".

To be honest, I need a hint how to make use of this hint.

(EDIT:) What I can do is compute:

$f(x) = (1 + \frac{1}{3}x) \big( (1+x) log(1+x) - x \big) \\ f'(x) = \frac{1}{3} \big( (1+x) log(1+x) - x \big) + (1 + \frac{1}{3}x)log(1+x) \\ f''(x) = \frac{(1 + \frac{1}{3}x)}{1+x} + \frac{2}{3}\log(1+\frac{1}{3}) \\ g(x) = \frac{1}{2}x^2\\ g'(x) = x \\ g''(x) = 1\\ \underset{x \rightarrow 0^+}{\lim}f(x)=\underset{x \rightarrow 0^+}{\lim}f'(x)=\underset{x \rightarrow 0^+}{\lim}g(x)=\underset{x \rightarrow 0^+}{\lim}g'(x)=0 \\ \underset{x \rightarrow 0^+}{\lim}f''(x) = 1 = \underset{x \rightarrow 0^+}{\lim}g''(x)$

Since all of functions above are continuous and differentiable on $(0,\infty)$, limits of $f$, $f$, $g$, $g'$ when $x \rightarrow 0^+$ are equal to $0$ and limits of $f''$ and $g''$ are finite and $g''(x) \ne 0$, we obtain: $$\underset{x \rightarrow 0^+}{\lim}(1 + \frac{1}{3}x)\, \mathcal{B}(x) = 1$$

But I don't understand the part of hint about "reducing the left hand side" to prove the inequality. Taking what I computed above and showing that the function is non-decreasing would be sufficient. But I want to understand method suggested by hint.

• Your function is going to have an imaginary component for $x \geq 1$. Have you copied it out correctly? – BadAtMaths May 13 '15 at 22:40
• @BadAtMaths Oh, thank you for spotting a mistake - I wrote minus instead of plus. I corrected it. – ltw May 13 '15 at 22:43

Let's see: $B(1) = 2 \ln 2 - 1= \ln (4/e)< \ln (4/(2.7)) = \ln (1 + (1.3)/(2.7))< \ln (1 +1/2)< 1/2.$ Now multiply by $(1+1/3) = 4/3.$ Is $(4/3)(1/2) > 1?$ Nope. Something's wrong here.
• if you compute the limite of $(1 + \frac{1}{3}x)\, \mathcal{B}(x)$ as $x$ tends to $0$ you will find $1/2$. So maybe one should prove that $(1 + \frac{1}{3}x)\, \mathcal{B}(x) \geq 1/2$ instead of $1$. – Idris May 14 '15 at 3:23
• the graphic by wolfram show $(1 + \frac{1}{3}x)\, \mathcal{B}(x) \geq 1/2 \, , \qquad x>0$ The function is increasing on (0, +\infty) and its limit at 0 is 1/2. – Idris May 14 '15 at 3:34
There is something strange in this problem, since, built at $x=0$, a Taylor expansion gives $$\mathcal{B}(x) = \frac{1}{x^2} \big( (1+x) \log(1+x) - x \big)=\frac{1}{2}-\frac{x}{6}+\frac{x^2}{12}-\frac{x^3}{20}+O\left(x^4\right)$$ Applying L'Hopital twice, you end with $$\frac{1}{2 (x+1)}$$ which just confirms the limit at $x=0$.
Then $$f(x)=(1 + \frac{x}{3})\, \mathcal{B}(x)=\frac{1}{2}+\frac{x^2}{36}-\frac{x^3}{45}+O\left(x^4\right)$$ Function $f(x)$ is continuously increasing and $f(x) >1$ only if $x>39.1242$.