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 April 25th, 2014, 11:36 PM #1 Senior Member   Joined: Apr 2008 Posts: 194 Thanks: 3 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.
 April 26th, 2014, 12:45 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra Yes, you can. Hint: for given two ration numbers $p=\frac{a}{b}$ and $q=\frac{c}{d}$ $$p \lt q \; \iff \; ad \lt bc$$
 April 26th, 2014, 03:28 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,809 Thanks: 2150 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?
 April 26th, 2014, 05:19 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,671 Thanks: 2651 Math Focus: Mainly analysis and algebra 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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