Prove two of $\frac{2}{a}+\frac{3}{b}+\frac{6}{c}\geq 6,\frac{2}{b}+\frac{3}{c}+\frac{6}{a}\geq 6,\frac{2}{c}+\frac{3}{a}+\frac{6}{b}\geq 6$ are True 
if $a,b,c$ are positive real numbers that $a+b+c\geq abc$, Prove that at least $2$ of following inequalities are true.
$\frac{2}{a}+\frac{3}{b}+\frac{6}{c}\geq 6, \space\space\space\space\frac{2}{b}+\frac{3}{c}+\frac{6}{a}\geq 6, \space\space\space\space\frac{2}{c}+\frac{3}{a}+\frac{6}{b}\geq 6$
Additional info: The Proof should be by contradiction.we can use Cauchy , AM-GM and other simple inequalities.

Things I have done so far:  I don't have a complete idea for this Problem.I just think
that for starting step I should prove at least one of those inequalities are true.
So, I assume that  $\frac{2}{a}+\frac{3}{b}+\frac{6}{c}< 6, \space\space\space\space\frac{2}{b}+\frac{3}{c}+\frac{6}{a}< 6, \space\space\space\space\frac{2}{c}+\frac{3}{a}+\frac{6}{b}<6$
.Summing these
inequalities gives us: $$\frac{11}{a}+\frac{11}{b}+\frac{11}{c}<18$$
using Cauchy and $a+b+c\geq abc$ I can write:$$ab+bc+ac\geq9$$
So I can rewrite Previous inequality as: $$\frac{99}{abc}<18$$
And I stuck here.
UPDATE
Thanks to user169478 help, we proved that at least one of these inequalities is true.So the remaining is to prove that if one of these 3 is true then the another one is true. So any hint for starting this part is appreciated.
As it was Proved that at least one of those 3 inequalities is true, I assume that $\frac{2}{a}+\frac{3}{b}+\frac{6}{c}\geq 6$ is true. Now we suppose that $\frac{2}{b}+\frac{3}{c}+\frac{6}{a}< 6$ and $\frac{2}{c}+\frac{3}{a}+\frac{6}{b}<6$.So we can say $$\frac{8}{b}+\frac{5}{c}+\frac{9}{a}<12$$
We have $\frac{2}{a}+\frac{3}{b}+\frac{6}{c}\geq 6$ So we can re write last inequality as $$\frac{7}{a}+\frac{5}{b}-\frac{1}{c}<6$$
and I stuck at proving this inequality is false.
 A: Set $x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}$.
So we have $xy+yz+xz\geq1$ due to $a+b+c\geq abc$ 
Assuming that all three inequalities are false, like you did, we obtain $11(x+y+z)<18$. But
$$(x+y+z)^2\geq3(xy+yz+xz)
\Rightarrow x+y+z\geq\sqrt{3}
\Rightarrow 11(x+y+z)\geq11\sqrt{3}>18$$
We have a contradiction.
Now we suppose only one of them is true. Assume that $2z+3x+6y\geq6$ is true. Then
$$2x+3y+6z<6
\Rightarrow4x^2+9y^2+36z^2+12xy+36yz+24xz<36 \tag{*}$$
Since $xy+yz+xz\geq1$ and using $*$ ,We obtain
$$4x^2+9y^2+36z^2+12xy+36yz+24xz<36(xy+yz+xz)$$
So
$$4x^2+9y^2+36z^2-24xy-12xz<0 \tag{1}$$
And using the same process, from $2y+3z+6x<6$ We obtain
$$4y^2+9z^2+36x^2-24yz-12xy<0 \tag{2}$$
Sum up $(1)$ and $(2)$ , We get
$$40x^2+13y^2+45z^2-36xy-12xz-24yz<0$$
which is equivalent to
$$9(2x-y)^2+4(y-3z)^2+(3z-2x)^2<0$$
We have a contradiction again. So we can conclude that at least two of the inequalities are true. 
A: This is a "brutal force" solution with the help of WolframAlpha. Hopefully someone else will give a more elegant proof later.
Set $x=\frac{1}{a}, y=\frac{1}{b}, z=\frac{1}{c}$, we have $xy+yz+xz\geq1$ due to $a+b+c\geq abc$
We can suppose, w.l.o.g, that $x\leq y \leq z$. Thus by the rearrangement inqeuality, $2x + 3y + 6z$ is the biggest one among the three sums. Since you have already proved that at least one of the three is no less than 6, we have $2x + 3y + 6z \geq 6$.
If $y \geq 1$, then we have $2z + 3x + 6y \geq 6$ for free, thus reach the conclusion.
Then we condiser the case when $x\leq y < 1$. Suppose $x= 1-a, y = 1-b$ for $0\leq b\leq a \leq 1$.
If $2y +3z + 6x < 6$ and $2z + 3x +6y < 6$, we have $$2 + 3z < 6a + 2b$$ $$2z + 3 < 3a + 6b$$ i.e.$$z < \frac{6a + 2b-2}{3}$$ $$z <\frac{3a + 6b-3}{2}$$.
Plug $x=1-a,y=1-b$ in $xy+yz+za \geq 1$, we get $z\geq\frac{a+b-ab}{2-a-b}$.
Thus we get $$\frac{6a + 2b-2}{3} > \frac{a+b-ab}{2-a-b}$$ $$\frac{3a + 6b-3}{2} > \frac{a+b-ab}{2-a-b}$$
i.e.$$6a^2 +2b^2+5ab -11a-3b+4 <0$$ $$3a^2+6b^2+7ab -7a-13b+6<0$$.
Draw these two curves in WolframAlpha, we get this.
Therefore we have the conclusion.
