One of the most fundamental inequalities in math is the Trivial Inequality, which says:
If x is a real number, then . if and only if .
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 for real numbers
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 , and the same is true of . The issue is that could be positive and bigger than the two squares, so we try again.
Moving the to the left side, we get a familiar looking expression, , which looks like it can be factorized. Sure enough, we get . 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 for real numbers
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 . Denominators are hard to work with, so we multiply by on both sides, to get . This is much more manageable and familiar too, as it looks like we can factorize it. Sure enough, we can factorize to get , which is again true by the Trivial Inequality.
Example 3 (a slightly trickier one):
Prove that for real numbers
We have squares and terms on both sides, so we move everything to one side. We get . We’re kind of at a loss now. We have squares, but we also have the subtracted terms. If only we had , so we could factorize…
We try this approach, subtracting from both sides and factorizing to get . This seems promising, as we have a nonnegative quantity now, but now we’re still stuck with the and . Also, as are not necessarily always positive, the right side could be positive as well, making things more tricky, so we try again.
Because we want and , we multiply both sides by two. . We try to factorize now, taking , to get . We don’t really see anything which indicates this won’t work, or will get messy and so we keep going, to get ! 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)
 Is true for real numbers where are of the same sign (if one is positive, the other is negative and vice versa)?
 Is the problem above true if are of opposite sign?
 Prove
 Prove (AMGM inequality in two variables)
 Find all solutions to the equation
 Find all integer solutions to the equation (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!
About the author

Zach Marinovhttps://etonstem.com/author/zach

Zach Marinovhttps://etonstem.com/author/zach

Zach Marinovhttps://etonstem.com/author/zach

Zach Marinovhttps://etonstem.com/author/zach
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