How to prove this lower bound on $\log(1+x)$ for $x \geq 0$? How about $x > -1$? I want to show that for all real numbers $x \geq 0$:
$$
\log(1+x) \geq \frac{x(5x+6)}{2(x+3)(x+1)}.
$$
I'd like to understand each step necessary to prove this so that I can apply it to future problems. Amazingly, a popular online calculator immediately spits out that this inequality is true for all $x > -1$. Is there an easy way to prove this for $x \geq 0$? How about $x >-1$?
Attempt (and more info on how I'm stuck):
My guess is to use the Taylor series expansion of both sides at $x=0$ (but why is that sufficient for all $x \geq 0$?). Expanding both sides:
$$
\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \geq x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{5x^4}{18} + \frac{7x^5}{27} - \ldots = \frac{x(5x+6)}{2(x+3)(x+1)}.
$$
I see that the early order terms match and then they don't. Not sure how to proceed from there. The signs are alternating on both sides, which I'm not sure how to deal with. Though, it seems the coefficients are converging to $0$ on both sides. It also seems the magnitude of the coefficients on the right-hand side is larger than the left-hand side (if you go term-wise), but how do I prove that? And do I need that fact to prove the inequality?
 A: Tbe problem looks complex  and encourages us to use approximations and taylor series but its infact very easy
Consider $$f(x)=\ln(1+x)-\frac{x(5x+6)}{2(x+3)(x+1)}$$ Indeed (after some hardwork)$$f'(x)=\frac{x^3}{{(x+3)}^2{(x+1)}^2}\ge 0$$
Thus $$f(x)\ge f(0)=0$$ Done
A: Basically, the same proof as the great @AlbusDumbledore except (1) it more closely follows @RobertShore's suggestion and (2) it's not as nice. It might be easier to follow step-by-step though.
Let $f(x) = \log(1+x)$ and $g(x) = \frac{x(5x + 6)}{2(x+3)(x+1)}$. We want to show that $f(x) \geq g(x)$ for all $x \geq 0$.
Note that $f(0) = g(0) = 0$. Thus, it suffices to show that $g^\prime(x) \leq f^\prime(x)$ for $x \geq 0$, i.e., that $f$ grows faster than $g$. Also, note that $f^\prime(x) = \frac{1}{1+x}$ and $g^\prime(x) = \frac{7x^2 + 15x + 9}{(x+3)^2(1+x)^2}$. Then,
\begin{align}
&\frac{7x^2 + 15x + 9}{(x+3)^2(1+x)^2} \leq \frac{1}{1+x} \\
&\iff \frac{7x^2 + 15x + 9}{(x+3)^2} \leq 1+x \\
%&\iff \frac{7x^2 + 15x + 9}{(x+3)^2} -1 \leq x \\
&\iff \frac{3x}{(x+3)}  \leq x \\
%&\iff \frac{3x}{(x+3)} - x  \leq 0 \\
&\iff \frac{-x^2}{(x+3)}  \leq 0 \\
&\iff \frac{x^2}{(x+3)}  \geq 0,
\end{align}
which is clearly true for $x \geq 0$.
