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 $\displaystyle a+1\geq 2\sqrt{a},b+1\geq 2\sqrt{b},c+1\geq 2\sqrt{c}$, we obtain

$\displaystyle a+b+c \geq \sqrt{a}+\sqrt{b}+\sqrt{c}+(\sqrt{a}+\sqrt{b}+\sqr t{c}-3)

$$\displaystyle \geq \sqrt{a}+\sqrt{b}+\sqrt{c}+(3\sqrt[3]{\sqrt{abc}}-3)

=\sqrt{a}+\sqrt{b}+\sqrt{c}$ - Jan 7th 2010, 02:42 PMLamalifInequalite and square root
a , b are positiv real such a+b=1

justify that:

$\displaystyle \sqrt{1+4b}+2\sqrt{1+4a}\geq 4$ - Jan 7th 2010, 03:20 PMDrexel28
- Jan 7th 2010, 07:13 PMShanks
Notice the fact that for any t in [0,1], we have

$\displaystyle \sqrt{1+4t}\geq 1+(\sqrt{5}-1)t$.

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

$\displaystyle

\sqrt{1+4a}+\sqrt{1+4b}\geq 4

$

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

$\displaystyle \sqrt{1+4a}+\sqrt{1+4b}\le 4$

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

$\displaystyle \Rightarrow a<1, b<1$

$\displaystyle \Rightarrow \sqrt{1+4a} <\sqrt{5},\sqrt{1+4b}<\sqrt{5}$

$\displaystyle \Rightarrow (\sqrt{1+4a})^2+2\sqrt{1+4a}\sqrt{1+4b}+(\sqrt{1+4 b})^2

<1+4a+2\sqrt{5}\sqrt{5}+1+4b=16$

$\displaystyle \Rightarrow \sqrt{1+4a}+\sqrt{1+4b}\le 4$ - Jan 11th 2010, 02:22 PMLamalifAnother answer
for any positive number n

$\displaystyle (1+2n)\le(1+n)^2$

$\displaystyle \sqrt{1+4a}\le\sqrt{(1+2a)^2},\sqrt{1+4b}\le\sqrt{ (1+2b)^2}$

$\displaystyle \sqrt{1+4a}+\sqrt{1+4b}\le(1+2a+1+2b)$