let a,b are positiv real such abc=1
prove that a+b+c>=sqrt(a)+sqrt(b)+ sqrt(c)
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}$
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.
since positive a and b satisify a+b=1,
$\displaystyle \Rightarrow a<1, b<1$
$\displaystyle \Rightarrow \sqrt{1+4a} <\sqrt{5},\sqrt{1+4b}<\sqrt{5}$
$\displaystyle \Rightarrow (\sqrt{1+4a})^2+2\sqrt{1+4a}\sqrt{1+4b}+(\sqrt{1+4 b})^2
<1+4a+2\sqrt{5}\sqrt{5}+1+4b=16$
$\displaystyle \Rightarrow \sqrt{1+4a}+\sqrt{1+4b}\le 4$
for any positive number n
$\displaystyle (1+2n)\le(1+n)^2$
$\displaystyle \sqrt{1+4a}\le\sqrt{(1+2a)^2},\sqrt{1+4b}\le\sqrt{ (1+2b)^2}$
$\displaystyle \sqrt{1+4a}+\sqrt{1+4b}\le(1+2a+1+2b)$