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

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 $\quad x+1-\dfrac{1}{2x^2}\le\frac32$  
