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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.

Sachin
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  • For (a) you can take $a=b=c=1/3$ for max, but min does not exist. Similarly max do not exist for (b) and (c). – Milly Oct 27 '14 at 17:28
  • Maybe of use: http://math.stackexchange.com/questions/991885/beautiful-little-inequality/991888#991888 – Kim Jong Un Oct 27 '14 at 17:34

3 Answers3

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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 $a=b=c=\frac13$.

Quang Hoang
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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).

Kim Jong Un
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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 $(1+1/a)(1+1/b)(1+1/c)\geq \frac{8}{\sqrt{abc}}\geq 8\sqrt{27}$

Dr. Sonnhard Graubner
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  • I don't think your approach for (c) works: to have equalities everywhere, you would have simultaneously $a=b=c=\frac{1}{3}$ and $a=b=c=1$. – Kim Jong Un Oct 27 '14 at 17:57