March | 2010 | Peeving Pedantry!

Problem: Prove that $a^2 + ab + b^2 \ge 3(a + b - 1)$ for all real $a, \, b$ .

The most natural approach is to consider the function $f(a, \, b) = a^2 + ab + b^2 - 3(a + b - 1) = a^2 +(b-3)a + (b^2 - 3b + 3)$ as a quadratic expression in , and prove that it is always non-negative using elementary properties of quadratic expressions and its discriminant. The problem can also be attacked through the Cauchy-Schwarz Inequality, and even elementary symmetric polynomials. But, these natural ways to attack this problem are not the motive behind this post. I want to share a very artificial way to attack this problem, and how very hard problems, can become trivial using this artificial approach. The artificial approach is nothing, but basic integration.

Solution: Due to symmetry in the variables, it is safe to assume that $b \ge a$ . We know use the well know inequality $x^2 + 1 \ge 2x$ . Next, we integrate both sides from to . And surprise! We obtain that

$\frac{(b^3 - a^3)}{3} + (b-a) \ge (b^2 - a^2)$

which is equivalent to $a^2 + ab +b^2 \ge 3(a + b -1)$ , since we assumed that $b \ge a$ , and we’re done.

This method can also be used to create, maybe hard, problems, as we can use any elementary inequality, integrate both sides, and get something unexpected. A few such examples are given below. All you need to do in order to prove them, is find an elementary inequality, and integrate both sides, and hope if turns out to be the inequality you wanted to prove!

Exercise 1: Prove that $a^4 + a^3 b + a^2 b^2 + a b^3 + b^4 + 15 \ge 10(a+b)$ for all real $a, \, b$ .

Exercise 2: Let $a_1, \, a_2, \, \dots , \, a_n$ be real numbers. Prove that

$\sum_{i=1}^{n} \sum_{j=1}^{n} \frac{a_i a_j}{i+j} \ge 0.$

Exercise 3: Let $a, \, b, \, c$ be positive real numbers. Prove that

$\frac{1}{3a} + \frac{1}{3b} + \frac{1}{3c} + \frac{3}{a+b+c} \ge \frac{1}{2a+b} + \frac{1}{2b+a} + \frac{1}{2b+c} + \frac{1}...$

Finally some not-so-trivial stuff for my blog.

Let denote the exponent of the prime number in the prime factorization of . If does not divide , then . I’ll leave you to prove the following, almost obvious, properties of .

$v_p(a+b) \ge \min \{ v_p(a), \, v_p(b) \}$

$v_p( \gcd (a_1, \, a_2, \, \dots , \, a_n) ) = \min \{ v_p(a_1), \, v_p(a_2), \, \dots , \, v_p(a_n) \}$

$v_p( \lcm (a_1, \, a_2, \, \dots , \, a_n) ) = \max \{ v_p(a_1), \, v_p(a_2), \, \dots , \, v_p(a_n) \}$

$v_p(n!) = \sum_{k \ge 1} \left\lfloor \frac{n}{p^k} \right\rfloor$

Note that $a \mid b$ if and only if for any prime we have $v_p(a) \le v_p(b)$ , and if and only if for every prime , .

Here’s a classical example to demonstrate the applications of :

Example: Prove the identity

$\frac{ \textrm{lcm} ( a, \, b, \, c)^2 }{ \textrm{lcm} (a, \, b) \cdot \textrm{lcm} (b, \, c) \cdot \textrm{lcm} (c, \,a )} =...$

for all positive integers .

Solution: Let , , and for an arbitrary prime . Now, in order to prove the identity, it suffices to show that the exponent of any prime in both the expressions is the same. Using the given introduction, this reduces to proving the fact that the two expressions

$2 \max \{x, \, y, \, z \} - \max \{x, \, y \} - \max \{y, \, z \} - \max \{z, \, x \}$ and

$2 \min \{x, \, y, \, z \} - \min\{x, \, y \} - \min \{y, \, z \} - \min \{z, \, x \}$ are in fact equal.

Due to the symmetry in the variables, it is safe to assume that $x \ge y \ge z$ . Now, it is easy to check that the two expressions are indeed equal, and we’re done.

Here are some more problems, which can also be solved using elementary properties of .

Problem 1: Let $a, \,b, \, c, \,d$ be positive integers such that . Prove that $\gcd(a, \, c) \cdot \gcd(a, \, d) = a \cdot \gcd (a, \, b, \,c, \,d)$ .