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.
Best Answer
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).