# A Simple but Nifty Inequality

(residuetheorem.com)

38 points | by vector_spaces 12 days ago

• cpp_frog 12 days ago
I saw this problem and it immediatly took me back to my undergrad days in which I had an optimal control class, where one of the topics studied is minimizing expressions of this kind by abstracting the problem to a tool called functional (, ) and then applying different methods of resolution, a combination of Lagrange multipliers and the Euler-Lagrange equation .

In this particular example, we take the expression to be minimized and then sum the restricions scaled by parameters λ1 and λ2. It should look like this (y=f(x)): ∫ y^2+λ1y+λ2xy dx integrated from 0 to 1. Then, we grab what is inside and apply the Euler-Lagrange equation: L(x,y)=y^2+λ1y+λ2xy ⇒ 2y+λ1y+λ2x=0. So now we now f(x) is linear (which is why the method in the post works, if f(x) was a polynomial of higher order, then it should have been substracted a poylynomial in x other than ax+b). Solving for y and substituting in the restrictions, we integrate linear and quadratic functions (easy) to arrive at a system of equations for λ1 and λ2. Solving we get (λ1,λ2)=(4,-12). Then f(x)=6x-2 is the minimizer for the functional. Integrating f(x)^2 we get that its integral is 4, so for any f(x) satisfying those two conditions, ∫ f(x)^2 dx is greater than or equal to 4, and that lower bound is achieved for f(x)=6x-2.

I didn't mention convexity, which is what allows us to conclude that the inequality is in that direction, but I leave it to the reader.

• Bobblebobble 12 days ago
We can also find the f minimizing the integral of f^2 by considering the Lagrangian: integral f(x)^2 - lambda f(x) + lambda - mu x f(x) + mu dx, then solving pointwise for f(x) to get f(x) = (lambda + mu x) / 2, then using the constraints to get lambda and mu.
• hansbo 12 days ago
I liked this because it uses a method to prove something which seems very disconnected from the initial problem.
• yobbo 12 days ago
It's not true if we pick a=0 and b=0?
• Smaug123 12 days ago
What's not true? The longest inequality on that page is true for all a, b; the rest of the article is about finding specific a, b to extremise the RHS. (Of course, since a=0 b=0 is not a critical point of the RHS, the subsequent lines are not true when a=0 b=0, since they're only true of a critical point.)
• yobbo 12 days ago
> The longest inequality on that page is true for all a, b

If you agree with this, then you why do also believe some particular a,b proves that ∫f²dx ≥ 4?

• cpp_frog 12 days ago
The premise is that any real number is non-negative when squared, so integration of a squared real function must yield a non-negative number. What is done next is that we now f(x)^2 is greater than a two-variable expression in a and b, so we try to find a maximum. It does not have to do with any particular values because of convexity or concavity of the two variable expression (e.g. if the minimum of a convex function is postivie, then the whole function is positive).
• yobbo 12 days ago
Yes. The maximum of the RHS is not the minimum of the LHS.
• Smaug123 12 days ago
Correct. The minimum of the LHS is at least the maximum of the RHS. This is more than you need.
• yobbo 12 days ago
What is the proven in the article is that the minimum of the LHS is at most 4 and at least 0.
• tgv 12 days ago
To expand on the other answer: by choosing a=0 and b=0, you show that the integral (the LHS) is greater than our equal to zero. The article then shows there are values that prove the lower limit is 4.
• yobbo 12 days ago
The article claims the RHS maximum is the minimum for the LHS.

> ... you show that the integral (the LHS) is greater than our equal to zero.

The article adds a function g, and uses (f+g)²≥0. It then picks a random g (g=ax+b), such that it can be simplified after expanding.

• Someone 12 days ago
> The article claims the RHS maximum is the minimum for the LHS.

No, it shows that the LHS is at least as large as the RHS for any reals a and b (1), then shows reals a and b exist that make the RHS equal to 4 (2), and from (1) and (2) concludes the LHS is at least as large as 4.

I don’t see it claim that that bound is tight, and the LHS can be made equal to 4.

• yobbo 11 days ago
It seems we have different interpretations of what ∫f²dx ≥ 4 means.

I read it as: no f exists for which ∫f²dx is less than 4.

• tgv 11 days ago
There might be a higher lower bound. So for all f that meet the requirements, ∫f(x)²dx ≥ d, the article shows that d is at least 4.

I can show that d is precisely 4, because that's the value for f(x) = 6x - 2.

• yobbo 11 days ago
f=6x-2 is just one example. You would need to show that no other f exists.