Prove that for every $x\in[0,1]$, this is true $\sqrt{1+2x}\geq x+1-(1/(2x^2))$ Prove that for every $x\in[0,1]$, this is true $\sqrt{1+2x}\geq x+1-(\frac{1}{2x^2})$
i proved that $x+1-\sqrt{1+2x}>0$ by: $(x+1)^2 -1-2x=x^2$ so $x+1>\sqrt{1+2x}$ but then don't know how to proceed for this question
Thank you in advance
 A: Added later:  I didn't initially see the pre-calculus tag.  I'm leaving the original answer as is, but appending a pre-calculus proof below it.  It turns out the pre-calculus proof is much much simpler!
One way to prove the inequality is to show that the function
$$f(x)=\sqrt{1+2x}-\left(x+1-{1\over2x^2}\right)$$
is strictly decreasing on the interval $(0,1)$, and then note that
$$f(1)=\sqrt3-{3\over2}\gt0$$
The derivative of $f$ is
$$f'(x)={1\over\sqrt{1+2x}}-\left(1+{1\over x^3}\right)$$
The term $1/\sqrt{1+2x}$ is never greater than $1$, and the term $(1+1/x^3)$ is never less than $1$, so the derivative is strictly negative, which implies the function is strictly decreasing.
Here's a precalculus proof:
For $0\lt x\lt1$, we have ${-1\over2}\gt{-1\over2x^2}$, so it suffices in fact to show that
$$\sqrt{1+2x}\ge x+1-{1\over2}$$
Both sides are now positive for all $x\in(0,1)$, and the right hand side simplifies to $x+{1\over2}$, so it suffices to show
$$1+2x\ge\left(x+{1\over2}\right)^2=x^2+x+{1\over4}$$
which reduces to showing that
$${3\over4}+x\ge x^2$$
But for $x\in(0,1)$, $x$ all by itself is greater than $x^2$, so the inequalities all hold.
A: For $0< x < 1/2$,  $x+1 - 1/2x^2<0$.
A: Each of the following are equivalent
$$
\begin{align}
\sqrt{1+2x}&\ge x+1-\frac{1}{2x^2}\tag{1}\\
\sqrt{1+2x}-(x+1)&\ge-\frac{1}{2x^2}\tag{2}\\
\frac{-x^2}{\sqrt{1+2x}+(x+1)}&\ge-\frac{1}{2x^2}\tag{3}\\[6pt]
\sqrt{1+2x}+(x+1)&\ge2x^4\tag{4}
\end{align}
$$
Reversible Operations:
$(2)$: subtract $x+1$ from both sides
$(3)$: rewrite the left side by multiplying by $\frac{\sqrt{1+2x}+(x+1)}{\sqrt{1+2x}+(x+1)}$
$(4)$: both sides are negative, so taking negative reciprocals preserves order
Now the left hand side of $(4)$ is $\ge2$ and the right hand side is $\le2$.

A far simpler approach 
Note that $\sqrt{1+2x}$ and $x+1-\dfrac{1}{2x^2}$ are increasing functions on $[0,1]$ and
on $[0,\frac58]$, $\quad\sqrt{1+2x}\ge1\quad$ while $\quad x+1-\dfrac{1}{2x^2}\le\frac{69}{200}\quad$ and
on $[\frac58\!,1]$, $\quad\sqrt{1+2x}\ge\frac32\quad$ while $\quad x+1-\dfrac{1}{2x^2}\le\frac32$  
