Sum of a Positive Number and its Reciprocal from Calculus

This is a short, animated visual proof demonstrating that sum of a positive real number and its reciprocal is always greater than or equal to 2.
To buy me a coffee, head over to www.buymeacoffee.com/VisualPr...
Thanks!
It turns out that this theorem is equivalent to the Arithmetic Mean-Geometric Mean inequality. The equivalence is implied by the following two proofs: • Visual Proof of AM-GM ...
• x + 1/x is greater tha...
And here are two alternate ways to prove this fact via the Pythagorean Theorem and triangle areas:
• x plus 1/x is greater ...
• x plus 1/x is greater ...
This animation is based on a proof by Roger B. Nelsen from the December 1994 issue of Mathematics Magazine, page 374 (doi.org/10.2307/2690999). You can find this here too: www.maa.org/sites/default/fil...
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[@sotocsick3195]

bro that's crazy. I never looked at functions like this

[@kmyc89]

I never was teached like this

[@ISuperI]

That's a creative use of calculus ngl

[@AzizKitten]

another method using calculus: compaire 1/x + x and 2 so we do 1/x + x - 2 = (x² - 2x + 1)/x = (x - 1)²/x (x-1)² always positive and we have x > 0 That means (x-1)²/x ≥ 0 1/x + x - 2 ≥ 0 1/x + x ≥ 2 but I liked your method

[@raymitchell9736]

I wished you'd give us a little more time at the end of the video to contemplate your question, but it helped to watch the video a second time.

[@DanielPoupko]

There's a nice visual proof for the AM-GM inequality, using the curveness of ln (or e^x, doesn't really matter) like here.

[@mr_flor]

wow, that's a way to prove it! so cool!

[@nbooth]

The function f(x) = x + 1/x is an example of a function that is invariant under inversion, meaning it is unchanged when x is replaced with 1/x. Another way of saying the same thing is that f(x) = x + 1/x is a solution to the functional equation f(x) = f(1/x). A function is invariant under inversion if and only if it is a solution to this functional equation. It is a kind of symmetry similar to functions that satisfy f(-x)=f(x), called "even" functions, which are symmetrical about the y-axis, except here the function on the interval (0, 1] is symmetric with the function on [1, ∞). There is a relationship between even functions and functions that are invariant under inversion. Hint: think about how log functions transform multiplication into addition.

[@khiemgom]

f(log(x)) is an even function?

[@jomilariola443]

This took me way to long, but I think I got the relationship? given an even function f(x) f(log(x)) is guaranteed to be invariant under inversion (took me multiple avenues/approaches on like guessing what the relationship could be, but landed me this, I wouldn’t be surprised if this isn’t the relationship) Edit: hmm, log can just be any function g where g(ab) = g(a) + g(b) and g and g(1) = 0 (implies g(x) = -g(1/x)) so given h(x) = f(g(x)), then h(x) = h(1/x)

[@konstantindrumev8036]

Hi, what text book would you recommend for studying that type of behaviours?

[@Amit_Pirate]

I was just learning about this function, KZbin algorithm works in mysterious ways

[@notohkae]

through your microphone

[@SHIVAPRASAD-hz4tj]

Such a beautiful proof.

[@luispinto2643]

Considering its given that x>0 couldnt you multiply both sides by x without any consequence?

[@Craftysob]

Beautiful

[@bigbadbith8422]

I have never seen this, and now I can’t not see it!😊

[@AbdoOpera]

Math is life ❤

[@violintegral]

Fun fact: the area bounded by the tangent line of f(x) = 1/x at any point x = a with the positive x and y axes is always 2.

[@clementfradin5391]

Another way to prove this : (x-1)^2 >= 0 (because a square is always positive) x^2-2x+1 >= 0 x^2+1 >= 2x Now divide by x, the inequality won’t change because x>0 x+1/x >= 2 The equality happens when x=1

[@richikdadhich9664]

Could also be proved using the property AM is greater than or equal to GM

[@RSLT]

COOL

[@sonictheone4568]

Hi! I’m curious, What software do you use to edit/create these videos?

[@MathVisualProofs]

These are all done in manim, which is the python library created by 3blue1brown

[@spegee5332]

yay jensen's

[@lugiagaurdien773]

1

[@user-LinaRexTone]

1 + 1/1 = 2, it's still true

[@lugiagaurdien773]

​@@user-LinaRexTone Yes, that is where equality holds.

[@ITS_MEEE333M]

▶️.

[@cornycornsnake]

No, I don’t see it. It seems to break as soon as you input x=2 or x=3. Am I missing something? If so, then what?

[@bradballinger4757]

When x=2: the function y=1/x passes through (2,1/2) while the line passes through (2,0). When x=3: curve goes through (3,1/3) while line goes through (3,-1). In both cases, the line passes through a lower point that the curve, which is to say that the line's y-value, 2-x, is less than the curve's y-value, 1/x. That is, 2-x

[@waarschijn]

Yeah I think I can see it 😉

[@kundansaurav2012]

|N+1/n|≥2 for n in R

[@aschetos]

x = 1

[@Antagon666]

So what ?

[@WG8PentelPen]

Let x > 0. f(x) = 1/x. This is a harded equation of a calculusi lmao

[@norahalhunayshil1762]

Whhaaat

[@RubyPiec]

this is the third proof you have of this

[@MathVisualProofs]

Fourth!

[@anonymus4339]

1

[@user-LinaRexTone]

1 + 1/1 = 2, it's still true