Could someone pls prove this inequality? (1 Viewer)

[Octavius]

 

[idkkdi]

View attachment 35102

xrx +1 >= x + rx
xrx - rx >= x -1
rx (x-1) >= x-1

 

[Octavius]

Oh I didn't realise that you were allowed to just move things around and prove. I thought you had to either use LHS - RHS or using the AM/GM inequality

 

[idkkdi]

Oh I didn't realise that you were allowed to just move things around and prove. I thought you had to either use LHS - RHS or using the AM/GM inequality

I would do this q the calculus way. LHS - RHS, then use derivatives and graphical reasoning to prove LHS - RHS >= 0 for x>= 0.

 

[Octavius]

Oh wait so I can't do what you did?

 

[idkkdi]

Oh wait so I can't do what you did?

you can move things around and prove, but what I did in the now deleted last step wasn't legit.

 

[Octavius]

so then would i write
rx >= 1
x>=0 which is true since given that x>=0?

 

[idkkdi]

so then would i write
rx >= 1
x>=0 which is true since given that x>=0?

rx (x-1) >= x-1,

we need to take two cases here, since x-1 might be negative and we would then get
rx<=1
which would still end in x>= 0.

So I guess it is valid haha.

 

[idkkdi]

rx (x-1) >= x-1,

we need to take two cases here, since x-1 might be negative and we would then get
rx<=1
which would still end in x>= 0.

So I guess it is valid haha.

I retract my following statement.

We would need to take three cases.
x-1 may be 0. In which case, we can't divide and we would need to check if x =1 works for the beginning statement, which it does.

 

[=)(=]

my solution

 

[idkkdi]

View attachment 35104
my solution

missing the 0<x<1 case

 

[=)(=]

ahh i see so it would be a negative times a negative still which is positive

Thanks for pointing it out though

 

[4321suomynona]

this is my solution
x√(x) + 1 >= x + √x
x√x - √x >= x - 1
√x (x-1) >= x-1
√x >= 1
x >= 1
(x-1) >= 0 (for x is real and x>1) and x-1 <= 0 (for x is real and x is between 0 and 1)

 

[idkkdi]

this is my solution
x√(x) + 1 >= x + √x
x√x - √x >= x - 1
√x (x-1) >= x-1
√x >= 1
x >= 1
(x-1) >= 0 (for x is real and x>1) and x-1 <= 0 (for x is real and x is between 0 and 1)

Need to look at cases for when you divide x-1 as said above.

 

[5uckerberg]

For inequalities, one of the good strategies would be to move everything onto one side and then factorise the problem.