Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

4 Answers 4

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 $x=1$).

share|improve this answer

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.

share|improve this answer

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 $x \ge 0$.

share|improve this answer

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.

share|improve this answer
    
Yes! Well done ;-) +1 –  amWhy Mar 7 '13 at 1:45

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.