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How can I prove the following inequality for $0

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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    The next step after what you provided will not work. – 2017-01-12
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    Very interesting question, Im stumpted too.. – 2017-01-12

5 Answers 5

4

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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    This is good. I just planned to add. :-) – 2017-01-12
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    @S.Y Thank you! Much appreciative. And Happy New Year! -Mark – 2017-01-12
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    Happy New Year to you too – 2017-01-12
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    @Dr.MV Thanks a lot for your answering. – 2017-01-12
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    @Susan20200 You're welcome! It was my pleasure. And Happy New Year! -Mark – 2017-01-12
3

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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    Thanks. But this is not an analytical proof that I am looking for. – 2017-01-11
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    @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 $f(x)=2^x$. – 2017-01-12
1

Consider the first derivate :$(\frac{\ln(1+x)}{\ln2})' = \frac{1}{\ln2(1+x)} > 0$ of $0

1

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 $1$.