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
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what do you mean sir? @Peter – lotofdots Jan 08 '14 at 19:28
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ye, i have it @Peter – lotofdots Jan 08 '14 at 19:38
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yes sorry i made mistake – lotofdots Jan 08 '14 at 19:42
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The statement is still false. – Alraxite Jan 08 '14 at 19:47
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okay i proved the following using your way: $0<\frac{1}{4x^4}+x^2-\frac{1}{x^2}-\frac{1}{x}$ but how to continue @Peter – lotofdots Jan 08 '14 at 19:48
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Thanks @Peter but it doesn't work either – lotofdots Jan 08 '14 at 19:59
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@Peter I retire what I said look at my edit – lotofdots Jan 08 '14 at 20:26
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@Alraxite Look at my edit – lotofdots Jan 08 '14 at 20:28
3 Answers
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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.
Barry Cipra
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For $0< x < 1/2$, $x+1 - 1/2x^2<0$.
Andrew
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user1942990
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0<x<1/2 --> x^2 + x^3 < 1/4 + 1/8 = 3/8<1/2 --> x^2 + x^3) < 1/2 --> 1 + x < 1/2(x2) --> 1/2, x+1−1/2x(2)<0 – user1942990 Jan 09 '14 at 20:08
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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 $\quad x+1-\dfrac{1}{2x^2}\le\frac32$
robjohn
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