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How can I prove the following inequality for $0<x<1$ $$\log_2{(1+x)} \geq {x}$$

I tried to use $\ln(1+x) \geq x-\frac{x^2}{2}$ for $x\geq 0$ and convert the $\ln$ to $\log_2$ to prove that, but it does not work. Any idea? Is there a generalized form?

I need an analytical proof, not sketch , or looking at graph. Thanks.

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  • $\begingroup$ The next step after what you provided will not work. $\endgroup$ – Susan_Math123 Jan 12 '17 at 0:00
  • $\begingroup$ Very interesting question, Im stumpted too.. $\endgroup$ – K Split X Jan 12 '17 at 0:02
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The logarithm function is concave.

Thus, it lies above any secant line.

In particular, on $[0,1]$, the function $x$ is a secant line for $\log_2(1+x)$.

Hence, $\log_2(1+x)\ge x$ on $[0,1]$.

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  • $\begingroup$ This is good. I just planned to add. :-) $\endgroup$ – S. Y Jan 12 '17 at 0:12
  • $\begingroup$ @S.Y Thank you! Much appreciative. And Happy New Year! -Mark $\endgroup$ – Mark Viola Jan 12 '17 at 0:13
  • $\begingroup$ Happy New Year to you too $\endgroup$ – S. Y Jan 12 '17 at 0:15
  • $\begingroup$ @Dr.MV Thanks a lot for your answering. $\endgroup$ – Susan_Math123 Jan 12 '17 at 2:32
  • $\begingroup$ @Susan20200 You're welcome! It was my pleasure. And Happy New Year! -Mark $\endgroup$ – Mark Viola Jan 12 '17 at 3:07
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Hint: Exponentiate the inequality to the equivalent form $1+x\ge2^x$, and sketch the two graphs $y=1+x$ and $y=2^x$. Note that $(x,y)=(0,1)$ and $(1,2)$ are common to both graphs.

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  • $\begingroup$ Thanks. But this is not an analytical proof that I am looking for. $\endgroup$ – Susan_Math123 Jan 11 '17 at 23:55
  • $\begingroup$ @Susan20200, this was only a hint. By looking at the two intersecting graphs, you should be able to see that the line lies above the curve because the curve is concave up. To prove that analytically, you need to take the second derivative of the function $f(x)=2^x$. $\endgroup$ – Barry Cipra Jan 12 '17 at 0:08
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Consider the first derivate :$(\frac{\ln(1+x)}{\ln2})' = \frac{1}{\ln2(1+x)} > 0$ of $0<x<1$, so our function is increase. $x$ also incease, so. We should just check what will be on one point. For example $x = 3$ second function is bigger, so on $0 < x < 1$ the first one will be bigger than second (because of they intersects in $x = 1$).

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As Barry said, convert to the equivalent form $1+x \geq 2^x$. Then, consider the function $f(x)=1+x-2^x$. Taking two derivatives you get that $f''(x)=-2^x (\ln(2))^2$ which is always negative. Hence $f$ is concave. Since $f(0)=0=f(1)$, it follows that $f$ must be positive on $(0,1)$, which is what you wanted.

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The key is, the derivative of the function $f(x)=\log_{2}(1+x)-x$ is $g(x)=\log_2 e\cdot\frac{1}{1+x} - 1$. The function $g(x)$ is decreasing for $x\in [0, 1]$, and the value is from positive to negative. So the minimum of $f(x)$ is at $0$ and $1$.

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