Math stuff and other things...but mostly math.: Inequalities and Optimization I

The following is a compilation of ideas and theorems on inequalities.

General Inequality:

The square of any real number will always be non-negative, or in other words,

$\displaystyle x^2 \geq 0$

for all real x.

Prove that for all real numbers x and y, that:

$(x^2 + 1)(y^2 + 1) \geq (xy + 1)^2$

Typically when you find your solution, you'll want to work backwards, that is, start with what you want to prove, and turn it into something you know is true. Then, when you present your solution, take what you know to be true, and use it to show what you want to prove. I'll present the above with the second method.

From our general inequality:

$(x-y)^2 \geq 0$

$x^2 - 2xy + y^2 \geq 0$

$x^2 - 2xy + y^2 + (xy)^2 + 1 \geq (xy)^2 + 1$

$x^2 + y^2 + (xy)^2 + 1 \geq (xy)^2 + 1 + 2xy$

Rearrane to make our factoring a bit more apparent:

$(xy)^2 + x^2 + y^2 + 1 \geq (xy)^2 + 2xy + 1$

This now factors into:

$(x^2 + 1)(y^2 + 1) \geq (xy + 1)^2$

Tadaa.

Arithmetic-Geometric (AM-GM) Inequality:

For positive numbers $a_1, a_2, a_3, \cdots , a_n$ , the arithmetic mean of them will be greater than or equal to their geometric mean, or in other words:

$\frac{a_1 + a_2 + a_3 + \cdots + a_n}{n} \geq \sqrt[n]{a_1a_2a_3\cdots a_n}$

Find the maximum value of:

$\frac{x}{x^2 + 3x+ 9}$

Call the maximum y. Rewriting this in terms of inequalities:

$\frac{x}{x^2 + 3x + 9} \leq y$

Invert both sides:

$\frac{x^2 + 3x + 9}{x} \geq \frac{1}{y}$

It's important to understand that when we invert both sides, the inequality sign is reversed. This is because the manual steps for inverting both sides ends up moving each term to the other side of the inequality.

Divide out the left side:

$x + 3 + \frac{9}{x} \geq \frac{1}{y}$

Rearrange the left side:

$x + \frac{9}{x} + 3 \geq \frac{1}{y}$

Applying AM-GM to our two x terms:

$\frac{x + \frac{9}{x}}{2} \geq \sqrt{(x)(\frac{9}{x})}$

The two x terms under the radical cancel out and turns into 9, and the entire RHS simplifies to 3.

$\frac{x + \frac{9}{x}}{2} \geq 3$

$x + \frac{9}{x} \geq 6$

Looking back at:

$x + \frac{9}{x} + 3 \geq \frac{1}{y}$

The only way we can minimize the left side is if we minimize the x terms. Since we've just determined the minimal value of them is 6, we can now write:

$6 + 3 \geq \frac{1}{y}$
$9 \geq \frac{1}{y}$
$\frac{1}{9} \leq y$

And there's our minimum, 1/9 .

However, there's an inconsistency in this solution (at least one which we have not addressed). We used AM-GM to find the maximum, but AM-GM only works for positive numbers. This means that we assumed x and 9/x are positive values. The reason we can do this without any errors is because our maximum will occur for a positive value of x. This can be confirmed by use of a graphing calculator (which is a big no-no on math competitions lawl), or algebraically.

When x is positive, the entire fraction is positive. When x is 0, the entire fraction is 0. When x is negative, the numerator will be 0, but since the minimum value of the quadratic in the numerator is positive (check the vertex), then the entire fraction will be negative. Therefore, the maximum value only occurs when x is positive.

The next post will be a follow up problem which uses these methods for solving it.