1. ## Math problem

let a,b are positiv real such abc=1
prove that a+b+c>=sqrt(a)+sqrt(b)+ sqrt(c)

2. 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)
$
$\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+(3\sqrt[3]{\sqrt{abc}}-3)
=\sqrt{a}+\sqrt{b}+\sqrt{c}$

3. ## Inequalite and square root

a , b are positiv real such a+b=1
justify that:
$\sqrt{1+4b}+2\sqrt{1+4a}\geq 4$

4. 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

5. 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!

6. ## Exact inequalite

a,b are positiv real such a+b=1
justify this
$
\sqrt{1+4a}+\sqrt{1+4b}\geq 4
$

thank you

7. 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.

8. 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

9. since positive a and b satisify a+b=1,
$\Rightarrow a<1, b<1$
$\Rightarrow \sqrt{1+4a} <\sqrt{5},\sqrt{1+4b}<\sqrt{5}$
$\Rightarrow (\sqrt{1+4a})^2+2\sqrt{1+4a}\sqrt{1+4b}+(\sqrt{1+4 b})^2
<1+4a+2\sqrt{5}\sqrt{5}+1+4b=16$

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

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