harder 3 Unit Inequalities question (2 Viewers)

[Hermes1]

it is given that for three positive real numbers a, b and c,


if we also no that a+b+c = 1, prove that:



new to inequalities, would someone be able to help me wiff this question.

and earlier in the question i proved:
- i dont no if this is relevant to above question.

 

[Drongoski]

This Q8 from Cambridge says: a, b, c > 0 and a+b+c = 1. Show that:

a) i) 1/a + 1/b + 1/c >= 9

ii) 1/(a^2) + 1/ (b^2) + 1/ (c^2) >= 27

b) ab + ac + ca >= 9abc

c) (1-a)(1-b)(1-c) >= 8abc

 

[Hermes1]

kk tnks ill try it.

 

[Drongoski]

a) (i)








(ii)







Now you carry on from here.

 

[Hermes1]

so with this topic u r just trying to manipulate expressions to suit the condition given?

 

[Omnipotence]

yeah

 

[Drongoski]

so with this topic u r just trying to manipulate expressions to suit the condition given?

You have to be very manipulative!

 

[largarithmic]

You dont need all that stuff listed earlier, you can do this question almost straight out using the AM-GM for two variables.



The first step is just using the condition, whereas the second step is just using .

 

[Hermes1]

You dont need all that stuff listed earlier, you can do this question almost straight out using the AM-GM for two variables.



The first step is just using the condition, whereas the second step is just using .

am i allowed to just use the AM-GM inequality without proving it?

 

[Omnipotence]

I went on google and saw Fermat's 'Last Theorem' and what is it?

 

[Hermes1]

I went on google and saw Fermat's 'Last Theorem' and what is it?

Fermat's last theorem states that no three positive integers a, b and c can satisfy the equation for any integer value of n greater that 2.

 

[Omnipotence]

I see.

 

[Drongoski]

You dont need all that stuff listed earlier, you can do this question almost straight out using the AM-GM for two variables.



The first step is just using the condition, whereas the second step is just using .

Brilliant. Didn't see that.

 

[Hermes1]

Drongoski am i allowed to just pull out this AM-GM thing without proving it?

 

[Drongoski]

Yes. One of the important basic inequalities for harder 3u.

 

[Hermes1]

Yes. One of the important basic inequalities for harder 3u.

nice. tnks for ur help.

 

[Drongoski]

I went on google and saw Fermat's 'Last Theorem' and what is it?

You are right to have cited in quotes "Last Theorem" because it should have been called Fermat's Last Conjecture. But because it has been referred to as such it has become known as "Fermat's Last Theorem". Why conjecture? Because until (now Sir) Andrew Wiles proved it in 1994/1995, over 350 years later, it was not established for sure whether that proposition was true or false.

 

[Omnipotence]

Didn't Andrew Wiles prove the Modularity Theorem (Taniyama-Shimura) rather than directly Fermat's?

 

[cutemouse]

it is given that for three positive real numbers a, b and c,


if we also no that a+b+c = 1, prove that:



new to inequalities, would someone be able to help me wiff this question.

and earlier in the question i proved:
- i dont no if this is relevant to above question.

Where did you get this q from?

 

[Hermes1]

Where did you get this q from?

2009 Shore Maths Extension 2 Trial Paper