If a,b,c $>0$ and a+b+c=1, then find the maximum / minimum value of the following If a,b,c $>0$ and a+b+c=1, then find the maximum / minimum value of the following : 
(a) abc 
(b) $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ 
(c) $(1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c})$ 
Using A.M - GM. inequality on a,b,c : 
$A.M \geq G.M
 $ 
$\frac{a+b+c}{3}\geq \sqrt[3]{abc}$ 
$\Rightarrow  \frac{1}{27} \geq abc $ 
If we use : let a =$\frac{1}{2},b=\frac{1}{3},c=\frac{1}{6}$ can we solve the inequalities by assuming these values somehow. please suggest thanks. 

for b) you can use AM-HM inequality namely
$\frac{a+b+c}{3}\geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$
this is equivalent to
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq9$
for c) you can use AM-GM we have
$1+1/a\geq \frac{2}{\sqrt{abc}}$ and etc we get by mulitplying
$(1+1/a)(1+1/b)(1+1/c)\geq \frac{8}{\sqrt{abc}}\geq 8\sqrt{27}$ 
