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I need to find a constant $a$ such that for all $x_1 x_2 > 0$:

$$a > - \frac{x_1^2 + 7 x_2^2}{2 x_1 x_2}$$

that is to say the supremum of the term on the right hand side. My question is how to find it. Of course, $x_1 x_2 > 0$ holds when both variables have the same sign. So I tried to simplify my problem a bit and now I want to find a constant $b = -2 a$ such that:

$$\frac{x_1}{x_2} + \frac{7 x_2}{x_1} > b$$

Looking at both inequalities, I thought 0 is the solution, but this can't be right. Because if $x_1$ is very small, the other term becomes very large, so there has to be a better boundary. Using some techniques (positive definite matrices), a friend of mine arrived at:

$$a > \sqrt{7}$$

however, I don't know how to apply those techniques yet and I have to get the result without those techniques. Does anyone have an idea, how I can get the boundary easily? I don't know whether $\sqrt{7}$ is correct by the way.

Thanks in advance.

PS: This is not a "real" homework, it's just something I think I have to show in order to solve the whole problem.

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1 Answer 1

up vote 3 down vote accepted

You want to minimize $\frac{x_1}{x_2} + \frac{7 x_2}{x_1}$. If you denote $t=\frac{x_1}{x_2}$, this is equivalent to minimizing $t+\frac7t$ for $t>0$.

This can be solved by using first and second derivative test. http://en.wikipedia.org/wiki/Maxima_and_minima

Alternatively, using AG-inequality http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means you get: $t+\frac7t \ge 2\sqrt{t\cdot\frac7t} = 2\sqrt 7$ and obviously, for $t=\sqrt 7$ the value $2\sqrt 7$ is attained.

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Thank you very much. –  Huy Apr 2 '11 at 12:10

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