Знаки добутків та тривіальна нерівність ($x^2\ge0$)

Inequalities (Part 1)

Main Inequality

For all real numbers $x$, $x^2 \ge 0$, with equality if and only if $x = 0$.

This is called the main inequality, and is the inequality that implies most inequalities. It may seem obvious, but it vital. This is already powerful enough to prove the AM-GM inequality for two variables :

For every positive real numbers $x, y$,

$\frac{x + y}{2} \ge \sqrt{xy}$.

Proof. Rewrite this as $(\sqrt{x} - \sqrt{y})^2 \ge 0$ and the results follows from the main inequality. Equality holds iff $x = y$.

The AM-GM inequality is a very useful inequality in general, and we'll cover that in more detail in the next part. But we continue to prove more inequalities with the main inequality. As an exercise, the reader is encouraged to try solving these problems. Hints will be given after pressing the Continue button.

Prove that

1. For all real numbers $a$, $a^2 \ge 6a - 9$.

2. For all positive real numbers $a, b, c$, $(a + b)(b + c)(c + a) \ge 8abc$.

3. For all real numbers $a, b, c$, $(a + b + c)^2 \ge 3(ab + bc + ca)$.

Hint for P1 : Move everything to the LHS and complete the square.

Hint for P2 : Use the 2-variable AM-GM inequality.

Hint for P3 : The inequality is equivalent to $2(a^2 + b^2 + c^2) \ge 2(ab + bc + ca)$ after multiplying both sides by $2$, then write it as the sum of three squares.

Press Continue for the solutions.

P1 : Rewrite as $(a - 3)^2 \ge 0$.

P2 : $a + b \ge 2\sqrt{ab}$. Multiplying similar inequalities gives the desired result.

P3 : Write it as $\frac{1}{2}[(a - b)^2 + (b - c)^2 + (c - a)^2]$.

This ends this part of the article.

Inequalities (Part 6)

This is not really a post on applications of famous inequalities to solve problems. This post will be on how to make an inequality problem (that might or might not be nice, depending on your luck) in under 5 minutes.

Step 1 : Pick a nonnegative term. (E.g. $(a + b - c)^2$)

Step 2 : Expand and try to factorise to make it as nice as possible, but not, of course, back to the original form.

Step 3 : After you got an inequality you like, you can add in AM-GM, Holder, Cauchy, or anything that seems approtiate to you.

You now produce an inequality problem! Repeat this process to produce more inequality problems.

Note : There are other ways to make inequality problems, this is just one of them.

The Trivial Inequality

The trivial inequality refers to the well-known fact that the square of a real number will always be greater than or equal to 0. You probably realized this sometime around when you first started learning how to square numbers, hence the name $\textit{trivial}$ inequality.

In math/numbers, we say that for any real number $x$, $$x^2 \geq 0.$$

You might be thinking, "how can something so... trivial be of any use?" The reality is that the trivial inequality is foundational in the study of inequalities. Let's start with an example problem...

Find all ordered pairs $(x, y, z)$ such that $$x^2 + y^2 + z^2 = 0.$$

Try applying the inequality...

$\textbf{Solution: }$ By the trivial inequality, $x^2 \geq 0, y^2 \geq 0, z^2 \geq 0$, and so $x^2+y^2+z^2 \geq 0$. For any real number $k$, the only time $k^2 = 0$ is when $k=0$, so clearly our only ordered pair is $(0,0,0)$. The case where $k=0$ is known as the $\textit{equality case}$ of the trivial inequality. Finding the equality case of any inequality is often pivotal in problem solving.

Now check out the following practice problems to deepen your understanding!

We'll consider 3 situations:

1. When $x>0$: If $x>0$, we have $x^2\ge0$ for all real values of $x$.

2. When $x=0$: Here, of course, $x^2=0$.

3. When $x<0$: If $x<0$, then $x^2>0$ again, because for any negative number $-a$, $$\left(-a\right)^2=\left(-a\right)\left(-a\right)=a^2,$$ which is clearly positive, or greater than $0$.

Thank you.