One of the most fundamental inequalities in math is the Trivial Inequality, which says:

If x is a real number, then x^2 \geq 0. x^2 = 0 if and only if x = 0.

This seemingly innocuous statement, which most of us may already know as ‘all squares are nonnegative’, can actually be used to prove much more complex inequalities. Let’s start with some examples. Also please note the equality case – that in order for a squared expression to be equal to 0, the expression itself must equal 0. This is a hint that it will be necessary in some of the exercises at the end.

Example 1:

Prove that a^2 + b^2 \geq 2ab for real numbers a, b

We see that there are some squares, which is what we want, however that’s not enough. All we can say about those is that a^2 \geq 0, and the same is true of b^2. The issue is that 2ab could be positive and bigger than the two squares, so we try again.

Moving the 2ab to the left side, we get a familiar looking expression, a^2 - 2ab + b^2, which looks like it can be factorized. Sure enough, we get (a-b)^2 \geq 0. Now, we have a square! And from the Trivial Inequality, it is true, as all squares are nonnegative.

In an actual proof, we would have to start with the factorized expression and work our way back, as this makes more logical sense, but if you’re not bothered about, then don’t worry.

Example 2:

Prove that b^2 / a^2 \geq 4b/a - 4 for real numbers a, b

This looks slightly trickier, because the algebra doesn’t look as nice. So let’s clean it up! Because our inequality has a square one side and 0 on the other, we try and replicate this by moving everything to the same side. Moving all of the terms to the left side, we get b^2 / a^2 - 4b/a + 4 \geq 0. Denominators are hard to work with, so we multiply by a^2 on both sides, to get b^2 -4ab + 4a^2 \geq 0. This is much more manageable and familiar too, as it looks like we can factorize it. Sure enough, we can factorize to get (b-2a)^2 \geq 0, which is again true by the Trivial Inequality.

Example 3 (a slightly trickier one):

Prove that a^2 + b^2 + c^2 \geq ab + bc + ca for real numbers a, b, c

We have squares and terms on both sides, so we move everything to one side. We get a^2 + b^2 + c^2 - ab - bc - ca \geq 0. We’re kind of at a loss now. We have squares, but we also have the subtracted terms. If only we had -2ab, so we could factorize…

We try this approach, subtracting -ab from both sides and factorizing to get (a-b)^2 + c^2 - bc - ca \geq -ab . This seems promising, as we have a nonnegative quantity now, but now we’re still stuck with the bc and ca. Also, as a, b, c are not necessarily always positive, the right side could be positive as well, making things more tricky, so we try again.

Because we want -2ab, -2bc and -2ac, we multiply both sides by two. 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac \geq 0. We try to factorize now, taking a^2 - 2ab + b^2, to get (a-b)^2 + a^2 + b^2 + 2c^2 - 2ac - 2bc \geq 0. We don’t really see anything which indicates this won’t work, or will get messy and so we keep going, to get (a-b)^2 + (a-c)^2 + (b-c)^2 \geq 0! And now we are done, as each of the squared expressions is nonnegative, so their sum is also nonnegative!

Note: for ‘is something true’ questions, either show it is not by providing a counterexample (a value which makes the inequality false) or prove that it is indeed true.

Exercises (all variables represent real numbers unless otherwise specified)

  1. Is a^2 + b^2 + ab \geq 0 true for real numbers a, b where a, b are of the same sign (if one is positive, the other is negative and vice versa)?
  2. Is the problem above true if a, b are of opposite sign?
  3. Prove a/b + b/a \geq 2
  4. Prove a/2 + b/2 \geq \sqrt{ab} (AM-GM inequality in two variables)
  5. Find all solutions to the equation 34y + z = 2\sqrt{34yz}
  6. Find all integer solutions x, y, z to the equation x^2 + 5y^2 + 10z^2 = 4xy + 6yz + 2z - 1 (AoPS)

When you’re done, email your answers to Zach Marinov, Paris Suksmith, or Aarit Bhattacharya for feedback, and don’t forget to post any questions in the comments section below!

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