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April 25th, 2014, 11:36 PM   #1
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How do you prove this inequality?

Here is the inequality I would like to ask about.

Show that sqrt(x^3 + 1)/(4x^3 + 2x + 1) < or = sqrt(x^3)/(4x^3).

I know that 1/(4x^3 + 2x + 1) < 1/(4x^3). I don't know how to show that the left numerator is less than the right denominator.

The website looks different from when I logged in last month.
Can you still do latex now?

Please help me with the math problem. Thanks.

Last edited by skipjack; April 26th, 2014 at 03:15 PM.
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April 26th, 2014, 12:45 AM   #2
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Yes, you can.

Hint: for given two ration numbers $p=\frac{a}{b}$ and $q=\frac{c}{d}$
$$p \lt q \; \iff \; ad \lt bc $$
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April 26th, 2014, 03:28 PM   #3
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Square both sides first.

To prove a/b < c/d (where a, b, c and d are positive), one can prove bc - ad > 0.

What polynomial does one get for bc - ad?
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April 26th, 2014, 05:19 PM   #4
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If it is
$$\sqrt{\frac{x^3 + 1}{4x^3 + 2x + 1}} \lte \sqrt{{x^3}{4x^3}}$$
there is no need to square both sides first. If the roots are of the numerators only, then you do need to square both sides (although you can do it after multiplying by both denominators).
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