If people asked you, what is the most elementary inequality you know? I bet your answer would be AM-GM. But in this series of posts I will try to show you the power that Cauchy Schwarz has over that of AM-GM. But as an introduction, let us first state and prove the theorem.
Cauchy Schwarz Inequality: Let (a1,a2,…,an) and (b1,b2,…,bn) be two sequences of real numbers, then we have:
(a12+a22+…+an2)(b12+b22+…+bn2)≥(a1b1+a2b2+…+anbn)2
In particular, equality holds iff there exists k∈R for which ai=kbi for i=1,…,n.
Proof: We will present 2 proofs, one originating from analysis on the equality case, the other by wishful thinking on small cases of n=2,3.
(i) Consider defining the following function f:
f(x)=(a1x−b1)2+(a2x−b2)2+…+(anx−bn)2.
We will expand this to get:
f(x)=(a12+a22+…+an2)x2−2(a1b1+a2b2+…anbn)x+(b12+b22+…+bn2)
From our first way of representing f(x), we can conclude that f(j)≥0↔∆f≤0 or
(a12+a22+…+an2)(b12+b22+…+bn2)≥(a1b1+a2b2+…+anbn)2
Equality holds if the equation f(x)=0 has one root.■
(ii) Just remark that:
(a12+a22+…+an2)(b12+b22+…+bn2)−(a1b1+a2b2+…+anbn)2=∑i,j=1n(aibj+ajbi)2≥0■
Let us note the following positive things regarding Cauchy-Schwarz:
Look forward to the next few posts to see applications of this extremely elegant inequality!
#Algebra
#Inequalities
#TorqueGroup
#KeyTechniques
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Hi everyone, I'm part of #TorqueGroup too. My assignment was to post olympiad level problems for the community that I found interesting. Since Anqi is posting these great topics about math things useful in such math, it seems like we're a perfect fit. So, here are some examples:
Show that for real numbers a1,a2,⋯ and b1,b2,⋯
i=1∑kbiai2≥∑i=1kbi(∑i=1kai)2
IMO 1995
Let a,b,c∈R+ such that abc=1. Prove that
a3(b+c)1+b3(a+c)1+c3(a+b)1≥23.
..
Hint: It may help to make a symmetric substitution for a=α,b=β,c=γ such that the condition αβγ=1 still holds.
USAMO 2009
For n≥2 let a1,a2,⋯,an be positive reals such that
(a1+a2+⋯+an)(a11+a21+⋯+an1)≤(n+21)2
Prove that max(a1,a2,⋯,an)≤4min(a1,a2,⋯,an).
..
Hint: The LHS should scream cauchy. Choose which ai multiplies which bk to isolate a desired max and min, and then do some algebra.
Log in to reply
I will be happy to post my solutions if requested (as long as you try them first!)
Could you add these examples to the Applications of Cauchy Schwarz Inequality Wiki page?
For the second one, following your hint, let a=x1,b=y1,c=z1. The expression is then ∑cycy+zx3yz. Since xyz=1, this is ∑cycy+zx2. Using Cauchy, this is greater than 2(x+y+z)(x+y+z)2. So now it is left to prove that x+y+z≥3. This is simple using AM-GM, so 3x+y+z≥3xyz, and we are done. ■
For the second one, WLOG assume a1 is the maximum and an is the minimum. Using Cauchy,
(a1+a2+⋯+an)(a11+a21+⋯+an1)≥(ana1+a1an+n−2)
(n+21)2≥(ana1+a1an+n−2)2
Taking square roots of both sides, we get
n+21≥ana1+a1an+n−2
25≥ana1+a1an
Now, squaring, we get
425≥ana1+2+a1an
Multiplying both sides by a1an we get
a12−417a1an+an2≤0
Factoring...
(a1−4an)(a1−41an)≤0
Since a1 and an are positive reals, (a1−4an) must be nonpositive. Thus,
(a1−4an≤0→a1≤4an, and we are done.
Log in to reply
Could you explain the second one as in how did you use Cauchy Schwartz in it.
Hope you share many more such techniques!!thnx a lot for this.
Log in to reply
Yup I will. In fact, I posted some on invariants and I also wrote some worked examples in my newest post. :)
What are symmetric inequalities?
One can easily prove Cauchy-Schwarz using vectors too. It is similar to the proof above, but is much cleaner. The interesting this is that almost every other inequality can be derived from this inequality.
Log in to reply
Cauchy-Schwartz is indeed a remarkable inequality. But there are many other inequalities that are considerably more subtle and/or general. The first of them is probably the Holder's inequality which is a direct generalization of the Cauchy-Schwartz.
@Anqi Li Can you add this to the Cauchy Schwarz Inequality Wiki page?