<s>[INST] Prove $\ln(1+x)\geq x-\frac{x^2}{2}$ 
When $x\geq0$ prove that: 
  $$\ln(1+x)\geq x-\frac{x^2}{2}.$$

My effort: From the Taylor series:
$$\ln(1+x)= x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$$
I don't know how to continue from here.
 [/INST] You want to prove $\ln(1+x)\ge x-\frac{x^2}{2}$ for all $x\ge 0$.
To prove it for $0\le x<1$, we'll use Taylor series. It converges because $-1<x\le 1$. See, e.g., here for more information. Or here.
$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$$
$\ln(1+x)\ge x-\frac{x^2}{2}$ is equivalent to $\frac{x^3}{3}-\frac{x^4}{4}+\cdots\ge 0$.
We'll prove for all $k\ge 3$, $k\in\mathbb Z$, $\frac{x^k}{k}-\frac{x^{k+1}}{k+1}\ge 0$. We could prove this using derivatives, but I'll use a simpler method here. The inequality is equivalent to
$\frac{x^k(k+1-kx)}{k(k+1)}\ge 0$
and we have
$\frac{k+1}{k}>1> x$, $x^k\ge 0$ because $x\ge 0$.
Using derivatives:
Proof: let $f(x)=(k+1)x^k-kx^{k+1}$.
Then $f'(x)=k(k+1)x^{k-1}(1-x)\ge 0$.
$f(x)\ge f(0)=0$.
Also notice that $\left(\frac{x^3}{3}-\frac{x^4}{4}\right)+\frac{x^5}{5}\ge 0$, where also $\frac{x^5}{5}\ge 0$, $\frac{x^{2t+1}}{2t+1}\ge 0$ for all $t\ge 2$, $t\in\mathbb Z$.
Therefore, if $0\le x<1$, then $\frac{x^3}{3}-\frac{x^4}{4}+\cdots \ge 0$, i.e. $\ln(1+x)\ge x-\frac{x^2}{2}$.
By the way, notice that when finding the sum $\frac{x^3}{3}-\frac{x^4}{4}+\cdots$, you have to find the limit of partial sums so that you have to sum the terms in the exact same order $\frac{x^3}{3}$, $\frac{x^3}{3}-\frac{x^4}{4}$, $\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}$, etc. because see the Riemann series theorem: summing the terms in a different order could result in a different number or diverge if the series is conditionally convergent. But in this case, $\frac{x^3}{3}-\frac{x^4}{4}+\cdots$ is not conditionally convergent when $-1<x<1$, because if $-1<x<1$, then it converges to $\ln(1+x)-x+\frac{x^2}{2}$ and $\frac{x^3}{3}+\frac{x^4}{4}+\cdots$ converges to $-\ln(1-x)-x-\frac{x^2}{2}$.
Here's a proof for $x\ge 1$. If $x>1$, then the Taylor series diverges, so we can't use it. $$x-\frac{x^2}{2}=\frac{2x-x^2}{2}=\frac{1}{2}-\frac{(x-1)^2}{2}\le $$
$$\le \frac{1}{2}<\ln 2=\ln(1+1)\le \ln(1+x)$$
because $1=\ln e<\ln 4=\ln 2^2$, i.e. $1<2\ln 2$, because $e<4$ and $\ln x$ is strictly increasing when $x>0$.
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