Proving a Quick and Easy Inequality (1+1/x)(1+1/y)≥9

If you need to post a picture of your solution or idea:
#ChallengingMathProblems #ProvingInequalities
EXPLORE 😎:

PLAYLISTS 🎵 :

[tangerin]

You can obtain the expression xy <= 1/4 by working with aritmetic mean >= geometric mean
(x+y)/2 >= sqrt(xy)
1/2 >= sqrt(xy)
xy <= 1/4

[shamanjitsingh]

I did it using calculus, I made a single function in x using the given equation and that function of x has minimum value of 9.

[drpkmath]

Hey there! Great job! This question reminds me of an old question I dealt with some years ago! Very smooth transitions!

[manojsurya]

Syber math's explanation are always clear, he mostly won't start directly with the solution and talks about how one might approach the problem and then explains them clearly, Ur great.and it was a great video

[marklevin]

Substitute 1-x for y. Then we have a function of x. Take derivative and it is negative for 0<x<1/2, positive for 1/2<x<1 and equals zero if x=1/2. Therefore function achieves its minimum value if x=1/2. And it's value is

[highlyeducatedtrucker]

Once you got to the 1 + (2/xy) >= 9 part, you can just multiply through by xy and do basic algebra manipulations to arrive at (1/4) >= xy. No need to invoke any identities.

[NoName-ehfz]

You can also use AM-GM for the 1/(xy) term and AM-HM for the 1/x +1/y term. You get that they are both <=4 :)

[brilliantbutlazy.]

You could have also used AM >= GM inequality to prove xy <= 1/4
(x+y)/2 >= (xy)^1/2
It directly follows after squaring and substituting x+y = 1 that xy <= 1/4 :)

[kumarsaurav]

Via homogenisation:
(x+1)(y+1)/xy>=9
Replace 1 by x+y to homegenise:
(2x+y)(2y+x)>=9xy
2x^2+2y^2-4xy>=0
2(x-y)^2>=0
which is trivially true
Equality when x = y = 1/2.

[echandler]

Again clear and easy to follow. IMO a great teacher who brings forth great comments.

[davidbrisbane]

I used Lagrange multipliers to find the minimum values of f(x, y) = (1+1/x)(1+1/y) subject to g(x, y) = x + y - 1 = 0, which results in a minimum of (1/2, 1/2), but in this case, it is easier to use the constraint to turn the problem into a single variable problem and solve that instead.

[adandap]

Neat - I used calculus to show that xy = x(1-x) <= 1/4 for all x in (0, 1). Your way is neater.

[mohamedihab]

After many simplifications and algebraic operations, I got the very convenient inequality: 4x^2 - 4x+1 >= 0, and the quadratic function has exactly one solution for when it is equal to zero (yes, it is when x=y=1/2) and since quadratic functions with positive x^2 coefficients are always increasing on both sides of the vertex point, therefore, the inequality is true, and we can use calculus directly with y= 1-x, but your proof (and many proofs here) are very elegant. Also great identity!

[advaykumar]

I knew AM -GM inequalities will be used befor the starting of the video

[AbouTaim-Lille]

Define : f(x) := (1+1/x) (1+1/(1-x) ) where x belongs to the open interval (0, 1). It is a well defined function since 1≠x≠0 . and f is differentiable at least twice. Notice that f(½) = 9 and f → ∞ whenever x →0, 1 . So it is enough to prove that f`(½) = 0, f is decreasing (ie f' negative) in the interval (0, ½) and f is increasing (ie f positive ) in the interval (½, 1).

[rcnayak_]

A little simplification. As x and y are real, (x - y) ^2 >= 0. From algebraic formula we know that the left hand side is also equal to (x + y)^2 - 4xy, so that (x + y)^2 - 4xy >= 0. That is 1 - 4xy >= 0 (as x + y = 1). That is 1 >= 4xy or 1/(xy) >= 4 that is the minimum value of 1/(xy) is 4. Therefore, the value of the expression 1 + (2 / (xy)) that we have already simplified, which can be written as 1 + 2 (1/(xy)) has the minimum value of 1 + 2.4 = 9.

[harshshukla]

Wow! Thanks for the upload, looking forward for another one tomorrow!

[tonyhaddad]

My way : first expand the inequality and substract 9 from both side then
Multiply both side by (xy) you get
X+Y+1-8XY >=0 since (x+y)=1 than
1+1-8XY >=0
2-8XY >=0
1-4XY >=0
(X=1-y) so 1 - 4(1-y)y >=0
4y^2 - 4y + 1 >=0
By the quadratic formula the only root is when y=1/2
Now if we take the derivitive you get
8y - 4 if we plug 1/2 into the derivitive we get 0 so the tangent lign at x = 1/2 is the x axis
Since this function is continous
Whether its always negative or always positive, now we have one step which is plug any number in this function finaly you get positive and we are done !!!! ■(proved)

[זאבגלברד]

I look at the function f(x)= 1 + 1/x and realise that the two x-values that are intresting for us are symetric to 1/2 so they are t and 1-t where t>0 but t<0.5 . and the product of the f - value of these two need to be greater thab 9. I put x=t and y=1-t and I got that the inequality stands for all t in the above range.

[MathElite]

Wow this is the first time I have seen this identity nice!