Prove that (1/x + 1/y + 1/z) is greater than 3

This was a pretty easy question but I showed 3 different ways of doing the proof. Feel free to play at faster speed.

[afuyeas]

If x, y, z are all positive one and add up to 1 they each have to be less than 1 but then their inverses are each bigger than 1 so 1/x + 1/y + 1/z > 1 + 1 + 1 > 3. That shouldn't take more than 20 seconds

[scottparkins]

It is actually quite easy to show that the inequality >= 9 (stronger than >3)…

Since x+y+z=1 it follows that

(1/x)+(1/y)+(1/z) = (x+y+z) *((1/x)+(1/y)+(1/z)).

Expanding this we get

(1/x)+(1/y)+(1/z) = 3 + (x/y) + (y/x) + (x/z) + (z/x) + (y/z) + (z/y).

Now for any a>0, it is easy to show that (1/a)+a >= 2 with equality iff a=1. Applying this to the above with a=x/y, x/z, y/z we get

(1/x)+(1/y)+(1/z) >= 3 + 2 + 2 + 2 = 9, with equality iff all numbers are equal to 1/3 .

Regardless I enjoyed the video since it gave a few different approaches - drives the point home that there isn’t always a unique approach 😊

[MadaraUchihaSecondRikudo]

Intuitively, the symmetric case is x=y=z=1/3, which would yield 9 in the second inequality which is larger than 3. Any other combination of x, y, z will require one of them to be less than 1/3 and another to be more than 1/3. The one that's less than 1/3 alone makes the sum of the reciprocals bigger than 3.

[NEKKITIS]

Simply substitute the sum x+y+z for the 1 in the numerator. You get:
x+y+z=1
1/x+1/y+1/z>3

Perform term-by-term division:

(y+z)/x+(x+z)/y+(x+y)/z>0
Because x, y, z > 0, the inequality is true.

[RyanLewis-Johnson-wqxs]

Mean Inequality Chain HM<GM<AM<QM HM=3/((1/x)+(1/y)+(1/z)), AM=(x+y+z)/3=1/3 HM ≤AM 3/((1/x)+(1/y)+(1/z))≤1/3 1/x+1/y+1/z ≥9 proves that 1/x+1/y+1/z >3

[martinmonath]

11:43 The nonmathematician inside me: c'mon, it's just a typo. The mathematician inside me: Please replace \alpha_n by \frac{1}{\alpha_n}.

[Jeremy-id]

Thank you for another wonderful and blessed video.

Substituting x + y + z for 1 in each term in 1/x + 1/y + 1/z is so cool and produces a great solution almost immediately.

I was familiar with the GM < or = AM inequality but had not come across those involving HM and QM before, so thank you for helping me to never stop learning.

Another proof using just the GM-AM inequality is as follows.

Let L = 1/x + 1/y + 1/z

Case 1: x=y=z
As in your solution, this means x=y=z=1/3 so L=3+3+3=9. Hence L>3.

Case 2: x y and z not all equal
Adding its 3 fractions,
L = S/xyz (1) where S = yz+zx+xy
By the GM-AM inequality,
S/3 > cube root (yz*zx*xy)
The inequality is strict because it is equal only when yz, zx and xy are all equal, which is not so in this case (because yz=zx means x=y and zx=xy means x=z, contradiction)
Hence S > 3 * cube root (x^2*y^2*z^2)
Observe that x, y and z are all > 0 and < 1. Hence x^2 > x^3 and similarly for y and z
Hence S > 3 * cube root (x^3*y^3*z^3)
= 3xyz = 3S/L from (1)
Hence S > 3S/L. And S>0. Hence L>3.

As others have noted, just 0 < x, y and z < 1 is enough to give 1/x, 1/y and 1/Z all > 1 and hence L>3. So the above proof is overkill but still fun for me to play with.

Again, thank you for your lovely videos and i look forward to seeing the next one. God bless you ❤

[ohiorizzler]

the actual sharp inequality is >=9, this is by cauchy

[TazBierzo]

Excellent !! I love seeing multiple approaches

[alexanderkhokhlov]

The simplest proof for me: Given that x + y + z = 1 and x, y, z > 0, we know that each of x, y, and z must be less than 1 because they sum up to one. Therefore, 1/x, 1/y, and 1/z are each greater than 1 because x, y, and z are less than one. Consequently, their sum is strictly greater than 3 because each term is more than one and there are three terms.

[stephanemoreau]

Another approach:
It is clear that x < 1, y < 1, and z < 1.
x < 1 implies that yz*x < yz (1)
y < 1 implies that xz*y < xz (2)
z < 1 implies that xy*z < xy (3)
Adding (1), (2), and (3) side by side yields:
3xyz < yz+xz+xy
Dividing both sides by xyz (which is positive) yields:
3 < (yz+xz+xy)/xyz = 1/x +1/y + 1/z Q.E.D.

[knowledgerrr]

Very nice!
First method is the simplest and more direct one.
Good to know about those terms (i.e. HM, GM and QM)
Thanks again

[kenjohnson]

Set it up as a constrained minimization problem using a Lagrange multiplier; you immediately get x = y = z.

[joansolerpascual]

What about doing this procedure:

Since x+y+z = 1 and x, y, z > 0 then:

a) x < x+y+z = 1 => x < 1 => 1 < 1/x
b) y < x+y+z = 1 => y < 1 => 1 < 1/y
c) z < x+y+z = 1 => z < 1 => 1 < 1/z

Then we can use that 3 inequalities to get our result:
1/x + 1/y + 1/z > 1 + 1 + 1 = 3

[doctorscoot]

if x+y+z=1 and are all positive real numbers, then assuming they are unequal, they will have some strict order x < y < z < 1. This implies that 1/x > 1/y > 1/z > 1, and therefore 1/x+1/y+1/z > 1+1+1, thus 1/x+1/y+1/z > 3. 🎉

[ghstmn]

Notice that x+1/x ≥2 for all x>0. The case obviously holds y, z so x+1/x + y+ 1/y z + z+ 1/z ≥6 => 1 + 1/x + 1/y + 1/z ≥ 5 > 3 hence proven

[Phylaetra]

Excellent presentation - I started by looking at minimizing with constraints, using calculus - but I really prefer all three of your methods!

[cristian.butacu]

We know that 1/x + x >= 2. The same idea applies for y and z. So we have:
1/x + x >= 2 (1)
1/y + y >= 2 (2)
1/z + z >= 2 (3)
From (1) + (2) + (3) we get that 1/x + 1/y + 1/z >= 5. Since 5 is greater than 3, the problem is solved.

[devashishjoshi]

Hello, nice video. One more method, Since, x>0, y>0, z>0, and x+y+z=1, so, 0<x<1, 0<y<1, 0<z<1, so 1/x>1, 1/y>1, 1/z>1, so, their sum is Strictly greater than 3.

[Devil-ncgb]

Thank you for making these high quality videos, you are helping me stay motivated to keep learning new maths.