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Show that $(\log(1+x))^2\le x$ and that $(\log(1+x))^2\le x^2$ for all $x\ge0$.

Both of these inequalities seem to be true, judging from plotting these functions with a grapher.

Can you help me get started? Can this be done by other means than with some painful Taylor series approaches?

Update. The latter inequality seems trivial, given $\log(1+x)\le x$, since we can just square both sides. So it seems like the first inequality is the non-trivial one.

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Are both inequalities supposed to be with the same expression on the left hand side? – Arturo Magidin Jun 25 '12 at 18:23
@ArturoMagidin Yeah. See my update though. – meh Jun 25 '12 at 18:23
Actually, your "update" suggests to me that the first inequality is meant to be $\log(1+x)\leq x$ rather than $(\log(1+x))^2\leq x$. Otherwise, the second inequality does not follow from the first for $0\leq x\lt 1$ by simple squaring. – Arturo Magidin Jun 25 '12 at 18:38
The first inequality is true for $x\geq 1$, I just proved it. – pritam Jun 25 '12 at 18:42
up vote 1 down vote accepted

For the first inequality, let $f(x)=\sqrt{x}-\ln(1+x)$. Then $f(0)=0$, and $$f'(x)=\frac{1}{2\sqrt{x}}-\frac{1}{1+x}=\frac{(1-\sqrt{x})^2}{\sqrt{x}(1+x)}\ge 0.$$ Thus $f(x)$ is an increasing function on $[0,\infty)$ (it hesitates a bit at $x=1$).

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I think the following Theorem can solve the first one directly:

Theorem: Let $f(x)$ and $g(x)$ be continuous in $[a,b]$ and differentiable in $(a,b)$, and suppose $f(a)=g(a)$. Moreover, suppose $f'(x)≥g'(x)$, where the equality does not hold identically in any subinterval $(\alpha,\beta)⊂[a,b]$. Then $f(x)>g(x)$ in $[a,b]$.

Take $g(x)=\log^2(x+1)$ and $f(x)=x$ on any interval $[0,b]$.

I hope my attempt can help.

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Yes! Well done ;-) +1 – amWhy Mar 7 '13 at 1:45

The first inequality holds for $x=0$, so it is enough to prove it for $x>0$.

For every $t>0$ we have $$ 0 \le (\sqrt{t}-1)^2=t-2\sqrt{t}+1 \iff 2\sqrt{t} \le 1+t \iff \frac{1}{1+t} \le \frac{1}{2\sqrt{t}}. $$ Integrating the latter inequality over $0<t\le x$ we get $\ln(1+x) \le \sqrt{x}$, i.e. $(\ln(1+x))^2 \le x$.

For the second inequality, notice that $$ \frac{1}{1+t} \le 1 \quad \forall t \ge 0. $$ Integrating the latter inequality over $0 \le t \le x$ we get $\ln(1+x) \le x$ for every $x \ge 0$, and since $u \mapsto u^2$ is increasing for $u \ge 0$, it follows that $(\ln(1+x))^2 \le x^2$ for every $x \ge 0$.

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Consider the Taylor series for $\log(1+x)$: $$\log(1+x) = x - \frac{x^2}{2}+\frac{x^3}{3} - \frac{x^4}{4}+\cdots + (-1)^{n+1}\frac{x^n}{n}+\cdots$$ which holds for $|x|\lt 1$ and $x=1$. So on $[0,1]$, this can be used to show $\log(1+x)\leq x$ (show that for such $x$, $\frac{x^{2k}}{2k} \geq \frac{x^{2k+1}}{2k+1}$, so that you are subtracting positive terms from $x$ in the computation).

But for $x\gt 1$ the approach does not work, since the Taylor series no longer converges. If you want to show that $\log(1+x)\leq x$, then you can instead show that $1+x \leq e^x$, by taking exponentials. Again you can use the Taylor series, this time for $e^x$: $$e^{x} = 1 + x + \frac{x^2}{2!} + \cdots + \frac{x^n}{n!} + \cdots$$ which makes $1+x\leq e^x$ for $x\geq 0$ immediate. This would establish the second of your inequalities as written at the current time, by squaring both sides of $\log(1+x)\leq x$ (since $\log(1+x)\geq 0$ when $x\geq 0$).

For $(\log(1+x))^2\leq x$, for $0\leq x\lt 1$ this follows from the first inequality by squaring, since $x^2\leq x$ on that region. For $x\gt 1$, a different approach would be required.

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