# Math problem

• Jan 5th 2010, 02:22 PM
Lamalif
Math problem
let a,b are positiv real such abc=1
prove that a+b+c>=sqrt(a)+sqrt(b)+ sqrt(c)
• Jan 6th 2010, 12:42 AM
Shanks
because $a+1\geq 2\sqrt{a},b+1\geq 2\sqrt{b},c+1\geq 2\sqrt{c}$, we obtain
$a+b+c \geq \sqrt{a}+\sqrt{b}+\sqrt{c}+(\sqrt{a}+\sqrt{b}+\sqr t{c}-3)
$
$\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+(3\sqrt[3]{\sqrt{abc}}-3)
=\sqrt{a}+\sqrt{b}+\sqrt{c}$
• Jan 7th 2010, 03:42 PM
Lamalif
Inequalite and square root
a , b are positiv real such a+b=1
justify that:
$\sqrt{1+4b}+2\sqrt{1+4a}\geq 4$
• Jan 7th 2010, 04:20 PM
Drexel28
Quote:

Originally Posted by Lamalif
a , b are positiv real such a+b=1
justify that:
[tex] \sqrt{1+4b}+2\sqrt{1+4a}\geq 4
[\math]

I meant, what do you want? Do you just want an answer? This is in the CALCULUS section. Do you want us to use Lagrange Multipliers? Do you want us to just use basic IMO like inequality stuff? You tell me
• Jan 7th 2010, 08:13 PM
Shanks
Notice the fact that for any t in [0,1], we have
$\sqrt{1+4t}\geq 1+(\sqrt{5}-1)t$.
Applying the fact, we get the inequality immediately!
• Jan 8th 2010, 10:27 AM
Lamalif
Exact inequalite
a,b are positiv real such a+b=1
justify this
$
\sqrt{1+4a}+\sqrt{1+4b}\geq 4
$

thank you
• Jan 8th 2010, 10:36 AM
Shanks
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, 04:11 PM
Lamalif
Excuse me (difficulty to use Latex)
the exact expression is this
a,b are positiv real such a+b=1
justify this

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

thank you again
• Jan 9th 2010, 03:07 AM
Shanks
since positive a and b satisify a+b=1,
$\Rightarrow a<1, b<1$
$\Rightarrow \sqrt{1+4a} <\sqrt{5},\sqrt{1+4b}<\sqrt{5}$
$\Rightarrow (\sqrt{1+4a})^2+2\sqrt{1+4a}\sqrt{1+4b}+(\sqrt{1+4 b})^2
<1+4a+2\sqrt{5}\sqrt{5}+1+4b=16$

$\Rightarrow \sqrt{1+4a}+\sqrt{1+4b}\le 4$
• Jan 11th 2010, 03:22 PM
Lamalif
$(1+2n)\le(1+n)^2$
$\sqrt{1+4a}\le\sqrt{(1+2a)^2},\sqrt{1+4b}\le\sqrt{ (1+2b)^2}$
$\sqrt{1+4a}+\sqrt{1+4b}\le(1+2a+1+2b)$