let a,b are positiv real such abc=1

prove that a+b+c>=sqrt(a)+sqrt(b)+ sqrt(c)

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- Jan 5th 2010, 01:22 PMLamalifMath problem
let a,b are positiv real such abc=1

prove that a+b+c>=sqrt(a)+sqrt(b)+ sqrt(c)

- Jan 5th 2010, 11:42 PMShanks
because , we obtain

- Jan 7th 2010, 02:42 PMLamalifInequalite and square root
a , b are positiv real such a+b=1

justify that:

- Jan 7th 2010, 03:20 PMDrexel28
- Jan 7th 2010, 07:13 PMShanks
Notice the fact that for any t in [0,1], we have

.

Applying the fact, we get the inequality immediately! - Jan 8th 2010, 09:27 AMLamalifExact inequalite
**a,b are positiv real such a+b=1**

justify this

thank you - Jan 8th 2010, 09:36 AMShanks
Is there a digit 2 in the problem?

If not, the inequality does not hold for some positive a,b, a+b=1.

If there is a digit 2, then applying the inequlity mentioned above, substitute t with a and b to get two inequality, add them up, the result follows immediately. - Jan 8th 2010, 03:11 PMLamalif
Excuse me (difficulty to use Latex)

the exact expression is this

**a,b are positiv real such a+b=1**

justify this

thank you again - Jan 9th 2010, 02:07 AMShanks
since positive a and b satisify a+b=1,

- Jan 11th 2010, 02:22 PMLamalifAnother answer
for any positive number n