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I am trying to prove that the following inequality holds for $ x > -1 $

$$\ln(1+x) - \frac{5x^2 + 6x}{2x^2 + 8x + 6} \ge 0$$

I've read on this site and others that a sufficient (but not necessary) condition to proving that this inequality holds is showing that the difference in first derivatives is non-negative. Trying that, I get:

$$\frac{1}{1+x} - \frac{7x^2 + 15x +9}{x^4 + 8x^3 + 22x^2 +24x + 9} = \frac{x^3}{(x^2 + 4x +3)^2}$$

For $x \ge 0$, this is clearly non-negative, but whenever $-1 < x <0$, the above expression is negative, which suggests that the approach using derivatives doesn't work. I graphed the difference for $-1 < x < 10$, and it does seem that the difference is always non-negative, but I'm not sure how to prove it for all $x>-1$. Any help would be appreciated.

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According to your calculations (with the derivative), this function has a local minimum at $x=0$ since it is decreasing in $(-1,0)$ and increasing in $(0,+\infty)$ But for $x=0$ $$\ln(1+0) - \frac{5\cdot0^2 + 6\cdot 0}{2\cdot0^2 + 8\cdot0 + 6}=0 \ge 0$$

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  • $\begingroup$ Wow, that's so simple. Totally overlooked the fact that this was true. Thanks! $\endgroup$ – Florian D'Souza Nov 9 '14 at 18:29
  • $\begingroup$ You are welcome. Practically you had it already by yourself! $\endgroup$ – Jimmy R. Nov 9 '14 at 19:28
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let $f(x)=\log(1+x)-\frac{5x^2+6x}{2x^2+8x+6}$ then we get $f'(x)=\frac{x^3}{(1+x)^2(3+x)^2}$ and we get $f'(x)=0$ for $x=0$ and further we obtain $f''(0)=0$ and $f'''(x)=0$ and $f^{(iv)}(0)=\frac{2}{3}\ne 0$ thus we have a minimum (global) for $x=0$

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Denote $f(x)=\ln(1+x)-\frac{5x^2+6x}{2x^2+8x+6}$, then $f'(x)=\frac{x^3}{(x^2+4x+3)^2}$ and $f(0)=0$, which imply that $f(x)$ decrease in $(-1,0)$ and increase in $(0,+\infty)$, so $f(x)$ attains its minimum at $x=0$, so $f(x)\geq0$.

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