<s>[INST] 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. 
 [/INST] As commented, max for (b) and (c) do not exist: simply let $a\to 0$. 
Min for (b) is achieved at $a=b=c=\frac13$ by AM-GM
$$\frac1a+\frac1b+\frac1c\ge 3\frac1{\sqrt[3]{abc}}\ge 9$$
and by (a).
To find min for (c) we use a trick:
$$1+\frac{1}{a}=1+\frac1{3a}+\frac{1}{3a}+\frac1{3a}\ge \frac{4}{\sqrt[4]{3^3a^3}}.$$
So
$$\left(1+\frac1a\right)\left(1+\frac1b\right)\left(1+\frac1c\right)\ge \frac{4^3}{\sqrt[4]{3^9(abc)^3}}\ge {4^3}.$$
The equality is attained at $a=b=c=\frac13$.
</s>