Lesson Inequalities (advanced)

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This Lesson (Inequalities (advanced)) was created by by richard1234(7193)

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About richard1234: BS in mathematics w/ computer science. Previous tutoring experience, and I have participated in the USAMO.

In the previous lesson, we learned the basics of inequalities. Here, we learn more advanced types of inequalities, including AM-GM, Cauchy-Schwarz, and in the next lesson, Muirhead's and the rearrangement inequality.

Perhaps the most basic inequality is the arithmetic mean - geometric mean (AM-GM) inequality, since many other inequalities are derived from this. It states that for any non-negative real numbers *[tex \LARGE a_1, a_2, ..., a_n], the arithmetic mean is greater than or equal to the geometric mean, i.e.

*[tex \LARGE \frac{a_1 + a_2 + ... + a_n}{n} \ge \sqrt[n]{a_1a_2...a_n}]

Equality occurs if and only if *[tex \LARGE a_1 = a_2 = ... = a_n]. It can be proven using induction, but I'll leave that up to you. Here, we look at a very common geometry problem:

---Problem---
A rectangular prism with dimensions l,w,h has a fixed surface area. Prove that the maximal volume occurs if and only if the rectangular prism is a cube.

---Solution---
We do NOT have to use Lagrange multipliers. Instead, we invoke AM-GM inequality. The surface area is given by S = 2(lw + wh + lh) and the volume is given by V = lwh. By AM-GM,

*[tex \LARGE \frac{lw + wh + lh}{3} \ge \sqrt[3]{l^2w^2h^2}]

*[tex \LARGE \frac{S}{6} \ge \sqrt[3]{V^2}]

The LHS is a constant, and from here it can be shown that the optimal solution is the "equality" case, where l = w = h, that is, the rectangular prism is a cube. &#8718;

AM-GM can actually be extended quite a bit. Define the generalized mean with power p, *[tex \LARGE M_p], as

*[tex \LARGE M_p = (\frac{1}{n} \sum_{i=1}^{n} a_i)^{\frac{1}{p}}]. Hence, if p = 1, we have the arithmetic mean, if p = -1, we have the harmonic mean. If p = 0, we assume it to be the geometric mean. The generalized AM-GM says that

If *[tex \LARGE p > q], then *[tex \LARGE M_p \ge M_q], equality holds when all *[tex \LARGE a_i] are equal.

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Another famous inequality is the Cauchy-Schwarz inequality, first published in 1821. It states that:

For any non-negative real numbers *[tex \LARGE a_1, a_2, ..., a_n] and *[tex \LARGE b_1, b_2, ..., b_n], the following inequality holds:

*[tex \LARGE (a_1^2 + a_2^2 + ... + a_n^2)(b_1^2 + ... + b_n^2) \ge (a_1b_1 + a_2b_2 + ... + a_nb_n)^2]. Equality holds if and only if a_i is a scalar multiple of b_i (that is, *[tex \LARGE a_i = kb_i] for all i).

A sample problem:

---Problem---
x,y,z are positive numbers. Prove that *[tex \LARGE x^2 + y^2 + z^2 - xy - yz - zx \ge 0].

---Solution---
The inequality we wish to prove is equivalent to *[tex \LARGE x^2 + y^2 + z^2 \ge xy + yz + zx]. By the Cauchy-Schwarz inequality,

*[tex \LARGE (x^2 + y^2 + z^2)(y^2 + z^2 + x^2) \ge (xy + yz + zx)^2]. Taking the square root of both sides,

*[tex \LARGE x^2 + y^2 + z^2 \ge xy + yz + zx]. &#8718;

There is another solution that works just as well:

---Alternate Solution---

Multiply both sides by 2:

*[tex \LARGE 2x^2 + 2y^2 + 2z^2 - 2xy - 2yz - 2zx \ge 0]. This factors quite nicely (you may check if you wish):

*[tex \LARGE (x - y)^2 + (y - z)^2 + (z - x)^2 \ge 0], which is obviously true since the square of any real number is at least zero, so the sum of three squares is at least zero. This is called the Trivial Inequality (named due to its triviality). &#8718;