# Proving inequalities using binomial expansion

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• Jun 22nd 2012, 10:32 AM
Kaloda
Proving inequalities using binomial expansion
If x and y are real nos. not equal to zero, show that
\frac{x^2}{y^2}+\frac{16y^2}{x^2}+24\geq\frac{8x}{ y}+\frac{32y}{x}.

The book hinted that binomial thingy (maybe expansion, I forgot) may be used.

EDIT:
BTW, please teach me how to use that \frac, \leq, etc.
• Jun 22nd 2012, 11:03 AM
Prove It
Re: Proving inequalities using binomial expansion
Quote:

Originally Posted by Kaloda
If x and y are real nos. not equal to zero, show that
\frac{x^2}{y^2}+\frac{16y^2}{x^2}+24\geq\frac{8x}{ y}+\frac{32y}{x}.

The book hinted that binomial thingy (maybe expansion, I forgot) may be used.

EDIT:
BTW, please teach me how to use that \frac, \leq, etc.

Your code looks alright, but you need to put it inside tex tags.
• Jun 22nd 2012, 02:46 PM
awkward
Re: Proving inequalities using binomial expansion
Quote:

Originally Posted by Kaloda
If x and y are real nos. not equal to zero, show that
$\frac{x^2}{y^2}+\frac{16y^2}{x^2}+24\geq\frac{8x}{ y}+\frac{32y}{x}$

The book hinted that binomial thingy (maybe expansion, I forgot) may be used.

EDIT:
BTW, please teach me how to use that \frac, \leq, etc.

Try multiplying both sides of the inequality by $x^2 y^2$ (this is positive, so you haven't changed the inequality).
Then rearrange all the terms on one side of the inequality (so you have $\text{bunch of stuff} \geq 0$) and see if you can apply the thingy.
• Jun 23rd 2012, 06:02 AM
Kaloda
Re: Proving inequalities using binomial expansion
Ah. Thank you for the proper syntax.

So this is the question
$\frac{x^2}{y^2}+\frac{16y^2}{x^2}+24\geq\frac{8x}{ y}+\frac{32y}{x}$

The book farther hinted that you may let $\frac{x}{y}=u$ then use the binomial expansion $(a+b)^4$
And BTW, if you know another solution, please post it even though it doesn't follow the book's solution.
• Jun 23rd 2012, 06:18 AM
Prove It
Re: Proving inequalities using binomial expansion
Well following the book's example we have

\displaystyle \begin{align*} \frac{x^2}{y^2} + \frac{16y^2}{x^2} + 24 &\geq \frac{8x}{y} + \frac{32y}{x} \\ \left(\frac{x}{y}\right)^2 + \frac{16}{\left(\frac{x}{y}\right)^2} + 24 &\geq 8\left(\frac{x}{y}\right) + \frac{32}{\frac{x}{y}} \\ u^2 + \frac{16}{u^2} + 24 &\geq 8u + \frac{32}{u} \\ u^4 + 16 + 24u^2 &\geq 8u^3 + 32u \\ u^4 - 8u^3 + 24u^2 - 32u + 16 &\geq 0 \\ 1(-2)^0 u^4 + 4(-2)^1 u^3 + 6(-2)^2 u^2 + 4(-2)^3 u^1 + 1(-2)^4 u^0 &\geq 0 \\ (u - 2)^4 &\geq 0 \end{align*}

which is a true statement and therefore confirms the original statement.
• Jun 23rd 2012, 08:14 AM
Kaloda
Re: Proving inequalities using binomial expansion
I already proved it by multiplying $x^2 y^2$ as what awkward said. But thanks Prove it for that brilliant solution.