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Are all inequalities based on the fact that a square is never negative?

Before any practice, I was once exposed to an olympiad problem that asked to prove a certain inequality. I had not a clue on how to tackle that problem. Then I found out about the AM-GM-HM inequalities, and I figured you would always have to manipulate those in order to prove a certain inequality. But surely these inequalities must be proved as well, so how is that done? I looked it up, and found out they're all based on the simple fact that a square of a real number is never negative.

So, my question is, are all inequalities based on that simple fact, in the end?

Note by Tim Vermeulen
4 years, 1 month ago

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Although a major number of them have the positivity of a square number at their core, it's wrong to say that all of them do. Some are based on the triangle inequality. Others, like Jensen's are based on the positivity of the second derivative. Udbhav Singh · 4 years, 1 month ago

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@Udbhav Singh This is a good comment, which raises more questions.

What is the triangle inequality based on? What makes a geometric picture true?

What is the positivity of the second derivative based on? Does this have anything to do with squares? Calvin Lin Staff · 4 years, 1 month ago

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@Calvin Lin As per my knowledge and understanding, triangle inequality could be easily proved by Pythagoras' theorem or cosine law. But obviously this method of proof only applies to euclidean geometry. Siddharth Kumar · 4 years, 1 month ago

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@Siddharth Kumar True, and i think there lies the answer... you cannot have a side's LENGTH to be negative.. Krishna Jha · 4 years, 1 month ago

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@Krishna Jha Indeed, that's a great observation! If you use the cosine rule, then it is based on the fact that \( -1 \leq \cos \theta \leq 1 \), which from unit circle trigonometry is based on the fact that if \( -1 \leq x \leq 1 \), then the distance of \((x,0) \) from \( (0,0) \) is also between -1 and 1.

This begs a further question. if points \(A, B, C \) are 3 points on a line in order, why must we have \( AB\leq AC \)? In other words, as you mentioned, why must the side lengths of a triangle be positive (non-negative)? Does this have anything to do with the trivial inequality? Calvin Lin Staff · 4 years, 1 month ago

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@Krishna Jha yes you are right. Siddharth Kumar · 4 years, 1 month ago

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Actually, I have found that most of the major ones are based on rearrangement. For example, most of the mean inquality chain is just cauchy which can be proved with rearrangement. Sandeep Silwal · 4 years, 1 month ago

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@Sandeep Silwal Cauchy can be proven by the Trivial Inequality as well, by considering the discriminant of the polynomial \[f(x)=\sum_{i=1}^n(a_ix+b_i)^2.\] David Altizio · 4 years, 1 month ago

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@David Altizio Interesting observation! Your next comment begs the question: If we prove something by only using discriminant considerations, does it avoid having to use the trivial inequality?

For example, we know that \( x^2 - \frac{1}{4} = 0 \) has real roots. Hence, the discriminant, which is \( 0^2 - 4 \times 1 \times ( -\frac{1}{4}) \) is greater than 0. So \(1>0 \). Does this use the trivial inequality anywhere? If so, how? Calvin Lin Staff · 4 years, 1 month ago

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@Calvin Lin Well, I know that the quadratic polynomial I wrote in my last post uses the Trivial Inequality in its discriminant considerations; since the polynomial in question is always greater than or equal to \(0\), its discriminant must always be less than or equal to \(0\). But I'm not sure if your polynomial uses the trivial inequality in the same way. David Altizio · 4 years, 1 month ago

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@David Altizio Your polynomial uses the trivial inequality in establishing that \(f(x) \geq 0 \). It then uses the discriminant condition of no real roots to prove Cauchy Schwarz.

Think about what the discriminant condition means, and if it requires the use of the trivial inequality. Calvin Lin Staff · 4 years, 1 month ago

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@Calvin Lin Well, the quadratic formula states that for any quadratic \(ax^2+bx+c=0\), we have \[x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},\] and the discriminant of this, \(b^2-4ac\), is inside a square root. Now, I think that the discriminant being less than zero implies nonreal roots, since otherwise there would be a real number \(r\) such that \(\sqrt{D}=r\implies D=r^2\), but since \(D<0\) this can not hold. So in essence the discriminant kinda has to do with the trivial inequality, since it follows directly in hand with the square root and its domain restrictions? David Altizio · 4 years, 1 month ago

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@David Altizio That is great analysis! At it's heart, the inequality from discriminants still arise from the trivial identity.

We have real roots if and only if the discriminant is a perfect square, which must be non-negative from the trivial inequality. Calvin Lin Staff · 4 years, 1 month ago

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@David Altizio Well yes :) Also, there are geometric inequalities such as R >= 2r and such which seem to be based upon the triangle inequality. Sandeep Silwal · 4 years, 1 month ago

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In addition to Jensen's, Bernoulli's inequality does not rely on this fact. The answer you're looking for though I guess is that it depends on how strong the inequality is. Remember that a certain type of problems can be solved with power means inequalities, another type requires Holder type inequalities( Cauchy Schwartz is a special case), other types fall for Schur or Muirhead. It may also be necessary to find suitable substitutions and play a lot with the different variables.In general weak inequalities will fall for Am-Gm and thus they are consequences of the trivial inequality. Jose R Urriola · 4 years, 1 month ago

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Actually, there are four basic inequalities that can imply the others: 1) A square is nonnegative 2) AM-GM 3) Cauchy-Schwarz 4) Rearrangement

So basically, if you assume any one of the four above as axiomatic, then you can prove the rest. There might be more, but those are the ones I remember. Bob Krueger · 4 years, 1 month ago

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@Bob Krueger I did not know about 3 and 4, but the AM-GM inequality can be proved by the fact that a square is nonnegative. It seems that the rearrangement inequality is not based on 1, though. :) Tim Vermeulen · 4 years, 1 month ago

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@Bob Krueger But 2,3,4 are all due to 1 Shourya Pandey · 4 years, 1 month ago

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@Shourya Pandey How do you get Rearrangement out of AM-GM? Udbhav Singh · 4 years, 1 month ago

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@Udbhav Singh No one suggests you can? Tim Vermeulen · 4 years, 1 month ago

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@Tim Vermeulen I was referring to the comment "But 2,3,4 are all due to 1" by Shourya P Udbhav Singh · 4 years, 1 month ago

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@Udbhav Singh 1 was "A square is nonnegative", not AM-GM Tim Vermeulen · 4 years, 1 month ago

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