Proving an Inequality Using Inequalities

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[SyberMath]

Hello everyone! Here is the link to Math Elite's channel! Check it out:

[michaelempeigne]

Quick method :

AM - HM inequality

( x + y + z ) / 3 > or = 3 / [ ( 1 / x ) + ( 1 / y ) + ( 1 / z ) ]

(x + y + z ) * [ ( 1 / x ) + ( 1 / y ) + ( 1 / z ) ] > or = 9

[dqrk]

am-hm proves this is 10 seconds
also with cauchy-schwartz it is pretty easy

[littlefermat]

This a straight forward application of the AM-GM inequality!
There was a saying we say a lot in math Olympiad "All inequalities can be proven using the AM-GM inequality in a suitable manner" 😁

[titassamanta]

Application of AM-GM inequality gives us (x+y+z) >= 3 × cube root of xyz and (1/x+1/y+1/z) >= 3 × cube root of 1/xyz. Multiply the inequalities to get the reqd result. (Note that x, y and z are positive and hence we could use the inequality.)

[lesnyk]

I blundered around with this for a bit and came to within a few yards of your #1 result, but got lost in the weeds! Thanks for taking me by the hand and leading me the rest of the way!

[davidbrisbane]

We can define
F(x, y, z) = (x+y+z)(1/x + 1/y + 1/z) and use multivariable calculus to find the minimum.

When we do this, we find the minimum occurs when
x² = yz, y² = xz and z² = xy.

_These equations tell us that_
_xyz = x³ = y³ = z³_

_So, x = y = z, when F(x, y, z) is a minimum, as x, y and z are all positive._

_So the minimum of_
_F(x, y, z)_
_= (x + x + x)(1/× + 1/x + 1/x)_
_= (3x)(3/x) = 9_

_So, F(x, y, z) ≥ 9, but let's go for overkill 😁_

If we substitute x² = yz into F(x, y, z), we find
F(x, y, z)
= (1 + y/x + z/x)² after several algebraic step
= (1 + y/x + x/y)², as x² = yz, so x/y = z/x

So, we need to show that (1 + y/x + x/y) ≥ 3.

We can now define G(x, y) = 1 + y/x + x/y and using multivariable calculus (again), we find the minimum occurs when x = y.

So, (1 + y/x + x/y) = 3 is the minimum of
G(x, y), as x = y.

So, F(x, y, z) ≥ (1 + y/x + x/y)² = 3² = 9.

Long story short. Multivarable calculus can solve this problem, but using inequalities is way easier and aesthetically more pleasing 😁

[goodplacetostop]

14:34 Can we extend it for 4 values x, y, z, t or even find a formula for n values : (x1 + x2 + … + xn)(1/x1 + 1/x2 + … + 1/xn) > X ?

[sabyasachimedda]

Replacing x with xy expresses the elegance of the solution.

[yoav]

I noticed that your second method is alwaus nicer than the first method, to make us watch the all video😀

[carlyet]

Symmetry x=y=z == 1 is the minimum as (1+ 1 + 1)(1+1+1) >= 9 at 9=9 if you perturb any variable the inequality increases. A.1/A = 1 so a 9 can be found 3.Ax 3.1/A == 9.1 :)

[krishnaats]

It was fun. I wrote lol in my notes looking at the long method..

AM>=HM will solve in a couple of seconds though..

[matniet]

The Cauchy - Schwartz inequality would be best here imo

[satyapalsingh]

Oooh, Very Interesting . Heart filled with joy

[ЧингизНабиев-эг]

Damn it, I remember getting this task on local olympiad and failing … :(

[otakurocklee]

Very nice. I was approaching the 2nd method but didn't figure it out.

[techysubham]

Best channel i have found so far, can we do beautiful questions of permutations and combinations

[ashikrasool]

Thanos: I used the inequality to prove the inequality

[tonyhaddad]

In my opinion the 2 methode are cool, 1st one help us to solve a more complicated inequality the 2nd one is less pain !!!

[gdtargetvn]

You can also use Sedrakyan's inequality to prove that 1/x+1/y+1/z is always greater or equal to 9/(x+y+z)
1/x+1/y+1/z >= ((1+1+1)^2)/(x+y+z)=9/(x+y+z) for any positive reals x, y, z
Therefore, for any positive reals x, y, z
The equality is only attained for x=y=z