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Problem

When studying Chernoff bound, one result is used without proof and reference, which is $$ (1+x)\log(1+x)-x\geq \frac{x^2}{\frac{2}{3}x+2} $$ I am wondering how this is proved.

What I Have Done

I checked the minimum of LHS and maximum of RHS, this indeed holds. But when it comes to prove this, this sort of check is far from enough.

Something I think relatable is doing some Taylor expansion of LHS, but I did not get the result.

Could someone help me, thank you in advance.

Edit

Take the second-order derivative of $f(x)=(1+x)\log(1+x)-x-\frac{x^2}{\frac{2}{3}x+2}$ gives us $f''(x)=\frac{x^2\, \left(x + 9\right)}{\left(x + 1\right)\, {\left(x + 3\right)}^3}$, which shows the correctness of the answer below.

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  • $\begingroup$ For which $x$ is the inequality to hold? For all $x>-1$? [need at least that restriction because of $\log(1+x)$] $\endgroup$
    – coffeemath
    Nov 18, 2018 at 3:20
  • $\begingroup$ Yeah, if you solve the inequality for the log, you get that it's $\ge \dfrac{5x^2+6x}{2x^2+8x+6}$ whose limit as $x \to \infty$ is $\dfrac 52$, and since $\log$ is unbounded its easy to see that after some point the inequality must be true. $\endgroup$
    – Ovi
    Nov 18, 2018 at 3:25
  • $\begingroup$ I don't think it is always true. Take x=9 for example. You get 1 on the left and 10.125 on the right. It is only true in (-1,0]. $\endgroup$
    – NoChance
    Nov 18, 2018 at 4:45
  • $\begingroup$ @NoChance. For $x=9$, $lhs=10\log(10)-9=14.0259$ $\endgroup$
    – Claude Leibovici
    Nov 18, 2018 at 4:49
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    $\begingroup$ The rhs is the $[2,1]$ Padé approximant (built at $x=0$) of the lhs. $\endgroup$
    – Claude Leibovici
    Nov 18, 2018 at 4:54

2 Answers 2

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Hint: Study the function $f(x) = LHS-RHS$ and show it is non-negative

Detailed hint: $f$ is infinitely differentiable on $(-1,\infty)$. Derive $f$ (twice): $f''$ is easy to handle, as it is a rational function (no more logarithms); it has a single root at $0$ and is always non-negative. This means $f'$ is non-decreasing; since $f'(0)=0$, we have $f$ decreasing on $(-1,0)$ and increasing on $(0,\infty)$. But $f(0)=0$, and thus $f(x)\geq f(0)=0$ for all $x$.

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    $\begingroup$ Thank you, but I think we can do this because we know RHS. I am wondering how something like RHS is derived from LHS using certain techniques. $\endgroup$
    – Mr.Robot
    Nov 18, 2018 at 5:04
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    $\begingroup$ Then you can for instance look at Taylor series to have an idea of a possible approximation by polynomials. For things like this one (not a polynomial), you would look at another type of approximation by the "best function in a simple class" (e.g., this: en.wikipedia.org/wiki/Pad%C3%A9_approximant) $\endgroup$
    – Clement C.
    Nov 18, 2018 at 6:04
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Also, we can make the following.

We need to prove that $f(x)\geq0,$ where $$f(x)=\ln(1+x)-\frac{5x^2+6x}{2(x^2+4x+3)}.$$ We see that $$f'(x)=\frac{x^3}{(x^2+4x+3)^2},$$ which says that $$f(x)\geq f(0)=0$$ and we are done!

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