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. 
 A: For (a), you can use the AM-GM inequality
$$
(abc)^{1/3}\leq\frac{1}{3}(a+b+c)=\frac{1}{3}\implies abc\leq\frac{1}{27}.
$$
For (b), use the AM-GM again, twice this time:
$$
(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq3\sqrt[3]{abc}\times3\frac{1}{\sqrt[3]{abc}}=9\implies\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 9.
$$
For (c), multiply out the LHS
$$
1+T_1+T_2+T_3\quad\text{where}\quad T_1=\frac{1}{abc},\\
T_2=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right),\quad T_3=\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right).
$$
Using (a) and (b), you have indeed estabished the minimum for $T_1$ and $T_2$ ($27$ and $9$, respectively). For $T_3$, you can use
$$
\frac{1}{3}T_3\geq(ab+bc+ca)\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\geq 9\implies T_3\geq 27.
$$
The first inequality is because $ab+bc+ca\leq\frac{1}{3}(a+b+c)^2$ (this one is true because it is equivalent to $(a-b)^2+(b-c)^2+(c-a)^2\geq 0$) and the second inequality above follows in the same manner as we have done in (b). 
A: 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 $a=b=c=\frac13$.
A: 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
$(1+1/a)(1+1/b)(1+1/c)\geq \frac{8}{\sqrt{abc}}\geq 8\sqrt{27}$ 
