# 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.

• For which $x$ is the inequality to hold? For all $x>-1$? [need at least that restriction because of $\log(1+x)$] – coffeemath Nov 18 '18 at 3:20
• 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. – Ovi Nov 18 '18 at 3:25
• 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]. – NoChance Nov 18 '18 at 4:45
• @NoChance. For $x=9$, $lhs=10\log(10)-9=14.0259$ – Claude Leibovici Nov 18 '18 at 4:49
• The rhs is the $[2,1]$ Padé approximant (built at $x=0$) of the lhs. – Claude Leibovici Nov 18 '18 at 4:54

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 $$x$$.
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!