Let $a,b,$ and $c$ be positive numbers with $a+b+c = 1$. Prove that $$\left (\dfrac{1}{a}+1 \right)\left (\dfrac{1}{b}+1 \right)\left (\dfrac{1}{c}+1 \right) \geq 64.$$


Expanding the LHS we obtain $\left (\dfrac{1+a}{a} \right)\left (\dfrac{1+b}{b} \right)\left (\dfrac{1+c}{c} \right)$. We are given that $a+b+c = 1$, so substituting that in we get $\left (\dfrac{b+c+2a}{a} \right)\left (\dfrac{a+c+2b}{b} \right)\left (\dfrac{a+b+2c}{c} \right)$. Then do I say $\left (\dfrac{b+c+2a}{a} \right)\left (\dfrac{a+c+2b}{b} \right)\left (\dfrac{a+b+2c}{c} \right) \geq 64$ and see if I can get a true statement from this?


8 Answers 8


Use the AM GM inequality : $$a+a+b+c\geq 4(a^2bc)^\frac{1}{4}$$ $$a+b+b+c\geq 4(ab^2c)^\frac{1}{4}$$ $$a+b+c+c\geq 4(abc^2)^\frac{1}{4}$$ Multiply the three inequalities and then divide $abc$ on both sides to get the desired inequality.


Sheer brute force is adequate:

$$(2a + b + c)(a + 2b + c)(a + b + 2c)$$

$$= 2(a^3 + b^3 + c^3) + 7(a^2b + ab^2 + a^2c+ac^2 + b^2c+bc^2) + 16(abc)$$

$$\geq 2(3abc) + 7(6abc) + 16(abc) = 64abc$$

  • $\begingroup$ Nice solution... $\endgroup$
    – qwr
    Dec 29, 2015 at 3:57

Probably, you already got the expected solution (from elementary algebra) in the other answers. So, I will give a different proof (from Multivariable Calculus). Specifically, I will use the Lagrange Multipliers Method to prove the following

Result: Given positive numbers $a$, $b$, $c$ and $d$ with $a+b+c=1$, we have $$\left(\frac{1}{a+d}+1\right)\left(\frac{1}{b+d}+1\right)\left(\frac{1}{c+d}+1\right)\geq\left(\frac{1}{\frac{1}{3}+d}+1\right)^3\tag{$*$}$$

Remark 1: Letting $d\to 0$ in $(*)$, we get $$\left(\frac{1}{a}+1\right)\left(\frac{1}{b}+1\right)\left(\frac{1}{c}+1\right)\geq\left(\frac{1}{\frac{1}{3}}+1\right)^3=64$$ as you want.

Remark 2: The same approach can be used to prove that for any positive numbers $x_1,...,x_n$ with $x_1+\cdots+x_n=c$ we have $$\left(\frac{1}{x_1}+1\right)\left(\frac{1}{x_2}+1\right)\cdots\left(\frac{1}{x_n}+1\right)\geq \left(\frac{n}{c}+1\right)^n$$ (See here the case $n=2$.)

Proof of Result: Let $d>0$ and $U=\{(a,b,c);\;a>0,\;b>0,\;c>0\}$. Define $f:U\to\mathbb{R}$ and $\varphi:U\to \mathbb{R}$ by $f(a,b,c)=\left(\frac{1}{a+d}+1\right)\left(\frac{1}{b+d}+1\right)\left(\frac{1}{c+d}+1\right)$ and $\varphi(a,b,c)=a+b+c$. Then, $$\nabla f(a,b,c)=-\left(\tfrac{1}{A^2}\left(\tfrac{1}{B}+1\right)\left(\tfrac{1}{C}+1\right),\tfrac{1}{B^2}\left(\tfrac{1}{A}+1\right)\left(\tfrac{1}{C}+1\right),\tfrac{1}{C^2}\left(\tfrac{1}{A}+1\right)\left(\tfrac{1}{B}+1\right)\right)$$ where $A=a+d$, $B=b+d$ and $C=c+d$, and $\nabla\varphi(a,b,c)=(1,1,1)$.

It follows from the Lagrange Multipliers Method that:

$$(a,b,c)\in \varphi^{-1}(1)\text{ is a critical point of } f\quad\Longleftrightarrow\quad\nabla f(a,b,c)=\lambda \nabla \varphi(a,b,c) \text{ for some } \lambda\in\mathbb{R}$$ $$\Longleftrightarrow \qquad\frac{1}{A^2}\left(\frac{1}{B}+1\right)\left(\frac{1}{C}+1\right) =\frac{1}{B^2}\left(\frac{1}{A}+1\right)\left(\frac{1}{C}+1\right) =\frac{1}{C^2}\left(\frac{1}{A}+1\right)\left(\frac{1}{B}+1\right)$$ $$\Longleftrightarrow\qquad \left\{ \begin{align}\frac{1}{A^2}\left(\frac{1}{B}+1\right) &=\frac{1}{B^2}\left(\frac{1}{A}+1\right)\\ \frac{1}{A^2}\left(\frac{1}{C}+1\right) &=\frac{1}{C^2}\left(\frac{1}{A}+1\right)\\ \frac{1}{B^2}\left(\frac{1}{C}+1\right) &=\frac{1}{C^2}\left(\frac{1}{B}+1\right)\end{align}\right.$$

$$\Longleftrightarrow\qquad\left\{\begin{align} B+B^2&=A+A^2\\ C+C^2&=A+A^2\\ C+C^2&=B+B^2 \end{align}\right.$$

$$\Longleftrightarrow\qquad\left\{\begin{align} B-A&=A^2-B^2=-(A+B)(B-A)\\ C-A&=A^2-C^2=-(A+C)(C-A)\\ C-B&=B^2-C^2=-(B+C)(C-B) \end{align}\right.$$

$$\Longleftrightarrow\quad A=B=C$$

$$\Longleftrightarrow\quad a=b=c=\frac{1}{3}$$

So, $q=(\frac{1}{3},\frac{1}{3},\frac{1}{3})$ is the unique critical point of $f$ in $\varphi^{-1}(1)$.

The expression of $f(a,b,c)$ defines a continuous function $g$ on $$\overline{\varphi^{-1}(1)}=\{(a,b,c);\;a\geq 0,\;b\geq 0,\;c\geq 0,\;a+b+c=1\}.$$ Since $\overline{\varphi^{-1}(1)}$ is compact, $g$ attains minimum at some point $p=(p_1,p_2,p_3)\in \overline{\varphi^{-1}(1)}$.

If $p\in \overline{\varphi^{-1}(1)}\setminus {\varphi^{-1}(1)}$ then $p_1=0$ or $p_2=0$ or $p_3=0$. Assume $p_1=0$. Then $$\left(\frac{1}{a+d}+1\right)\left(\frac{1}{b+d}+1\right)\left(\frac{1}{c+d}+1\right)\geq \left(\frac{1}{d}+1\right)\left(\frac{1}{p_2+d}+1\right)\left(\frac{1}{p_3+d}+1\right)$$ which is a contradiction (because letting $d\to 0$, the LHS goes to some constant and the RHS goes to $\infty$). It follows that $p_1= 0$ is impossible. The same argument shows that $p_2\neq 0$ and $p_3\neq0$. Therefore, $p\in\varphi^{-1}(1)$. Since $g=f$ on $\varphi^{-1}(1)$, we conclude that $f$ attains minimum in $\varphi^{-1}(1)$. Thus, the minimum value of $f$ in $\varphi^{-1}(1)$ is $f(q)$ (because $q$ is the unique critical point of $f$ in $\varphi^{-1}(1)$).

Therefore, for all $(a,b,c)\in\varphi^{-1}(1)$ we have $$f(a,b,c)\geq f\left(\tfrac{1}{3},\tfrac{1}{3},\tfrac{1}{3}\right)\tag{#}$$ As $(\text{#})$ is $(*)$, the proof is complete. $\Box $


By AM-GM we have:

$1 + \frac{1}{a} = \frac{1}{a}(a + b + c + a) \ge \frac{1}{a}4\sqrt[4]{{{a^2}bc}}$

$\Rightarrow 1 + \frac{1}{a} \ge \frac{4}{a}\sqrt[4]{{\frac{{{a^4}bc}}{{{a^2}}}}} = 4\sqrt[4]{{\frac{{bc}}{{{a^2}}}}} $

And $1 + \frac{1}{b} \ge 4\sqrt[4]{{\frac{{ca}}{{{b^2}}}}};1 + \frac{1}{c} \ge 4\sqrt[4]{{\frac{{ab}}{{{c^2}}}}}$

$$\Rightarrow \left( {1 + \frac{1}{a}} \right)\left( {1 + \frac{1}{b}} \right)\left( {1 + \frac{1}{c}} \right) \ge 4\sqrt[4]{{\frac{{bc}}{{{a^2}}}}}4\sqrt[4]{{\frac{{ca}}{{{b^2}}}}}4\sqrt[4]{{\frac{{ab}}{{{c^2}}}}} = 64$$


Here is a "simple" proof utilizing GM-HM inequality.

Use GM-HM inequality on the set $\{\frac{a+1}{a},\frac{b+1}{b},\frac{c+1}{c}\} $ to get:

$$\left(\,\left(\frac{a+1}{a}\right)\left(\frac{b+1}{b}\right)\left(\frac{c+1}{c}\right)\,\right)^\frac{1}{3} \geq \frac {3} {\left(\frac{a}{a+1}\right) +\left(\frac{b}{b+1}\right) + \left(\frac{c}{c+1}\right)} $$

$$\implies \,\left(\frac{a+1}{a}\right)\left(\frac{b+1}{b}\right)\left(\frac{c+1}{c}\right) \geq \frac {27} {\left( \,\, \left(\frac{a}{a+1}\right) +\left(\frac{b}{b+1}\right) + \left(\frac{c}{c+1}\right)\,\,\right)^{3}} \,\,\,(♦)$$

Now, maximizing $\left(\frac{a}{a+1}\right) +\left(\frac{b}{b+1}\right) + \left(\frac{c}{c+1}\right)$ according to the constraint $a+b+c=1$ , its maximum value comes out to be $\frac{3}{4}$.

$$\therefore \left(\frac{a}{a+1}\right) +\left(\frac{b}{b+1}\right) + \left(\frac{c}{c+1}\right) \leq \frac {3}{4}\,\,\,(♣)$$

Plug $(♣)$ in $(♦)$ to get :

$$\left(\frac {a+1}{a}\right) \left( \frac{b+1}{b}\right) \left(\frac{c+1}{c}\right) \geq \frac {27} {(3/4)^3} =\frac {27\times 64}{27} = 64$$


$$\left(\frac{1}{a}+1\right)\left(\frac{1}{b}+1\right)\left(\frac{1}{c}+1\right) \geq 64$$

as desired.

$$HENCE \,\,\,\, PROVED$$

  • $\begingroup$ $$\implies \,\big(\frac{a+1}{a}\big)\big(\frac{b+1}{b}\big)\big(\frac{c+1}{c}\big) \geq \frac {27} {\big( \,\, \big(\frac{a}{a+1}\big) +\big(\frac{b}{b+1}\big) + \big(\frac{c}{c+1}\big)\,\,\big)^\color{red}{{\frac{3}{1}}}} \,\,\,(♦)$$ Is this what you mean or I missed something ? $\endgroup$
    – user312097
    Feb 4, 2017 at 8:03
  • $\begingroup$ @A---B O yeah you are right.... just a typo .... edit on its way. Thanks for correcting me. $\endgroup$
    – user399078
    Feb 4, 2017 at 8:14
  • $\begingroup$ @A---B How do you rate my answer ???? Is it ok or should I add fine-tune it further ... ???? $\endgroup$
    – user399078
    Feb 4, 2017 at 8:15
  • $\begingroup$ It is good but if you have time you can use \left and \right to make brackets of apropriate size. like $({3 \over 2})$ becomes $\left({3 \over 2}\right)$. $\left({3 \over 2}\right)$ $\endgroup$
    – user312097
    Feb 4, 2017 at 12:48
  • $\begingroup$ @A---B Thanks for the suggestion ! :) $\endgroup$
    – user399078
    Feb 4, 2017 at 12:54

Probably you mean $a,b,c\geq 0$. You want to find the miminmum of $$\ln (1+\frac{1}{a})+\ln(1+\frac{1}{b})+\ln(1+\frac{1}{1-a-b})$$ for $a,b\geq 0$. The partial derivatives are zero iff $$ -\frac{1}{a(a+1)}+\frac{1}{(1-a-b)(2-a-b)}=0 $$ and a similar equation in which $a$ and $b$ will be replaced. By subtracting, you get $a(a+1)=b(b+1)$ which implies (given that all is positive) $a=b$. But that's the minimum, $a=b=c=1/3$.


In fact we can even prove a general form of this inequality. $$\left(1+\frac{1}{x_1}\right)\left(1+\frac{1}{x_2}\right)...\left(1+\frac{1}{x_n}\right)\leqslant(1+n)^n(*)$$

Where: $x_1, x_2, ..., x_n\gt0$ and $x_1+x_2+...+x_n=1$.


Firstly let's consider this matrix

\begin{pmatrix} a_{11} & a_{12} & ... & a_{1n}\\ a_{21} & a_{22} & ... & a_{2n}\\ . & . & . & .\\ . & . & . & .\\ a_{m1} & a_{m2} & ... & a_{mn} \end{pmatrix}

And prove the following lemma:

$$A(G_1, G_2, ..., G_m)\leqslant G(A_1, A_2, ..., A_n)(**)$$ where $A(x_1, x_2, ..., x_n)=\frac{x_1+x_2+...+x_n}{n}$,

$G(x_1, x_2, ..., x_n)=\sqrt[n]{x_1x_2...x_n}$,

$A_i$ is an arithmetic mean of i-th column of the matrix, $i={1,2,...,n}$.

And $G_j$ is a geometric mean of j-th row of the matrix, $j={1,2,...,m}$.

If any of $A_i=0$ then $(**)$ is obviously true, so let $A_i \gt 0$ then by the Cauchy–Schwarz inequality we have:$$\sum_{j=1}^n \frac{a_{ij}}{A_j}\geqslant\frac{nG_i}{G(A_1, A_2, ..., A_n)}$$

from ths we get $$\frac{mA_1}{A_2}+\frac{mA_2}{A_2}+...+\frac{mA_n}{A_n}\geqslant\frac{nmA(G_1, G_2, ..., G_m)}{G(A_1, A_2, ..., A_n)}$$ so finally $$A(G_1, G_2, ..., G_m)\leqslant G(A_1, A_2, ..., A_n)$$ which proves our lemma.

Now let's construct following matrix \begin{pmatrix} 1 & 1 & ... & 1\\ \frac{1}{x_1} & \frac{1}{x_2} & ... & \frac{1}{x_n}\\ \end{pmatrix}


$G(A_1, A_2, ..., A_n)=G(\frac{1+\frac{1}{x_1}}{2}, \frac{1+\frac{1}{x_2}}{2}, ..., \frac{1+\frac{1}{x_n}}{2})=\frac12\sqrt[n]{(1+\frac{1}{x_1})(1+\frac{1}{x_2})...(1+\frac{1}{x_n})}$,

$A(G_1, G_2)=A(1, \sqrt[n]{\frac{1}{x_1x_2...x_n}})=\frac12(1+\frac{1}{\sqrt[n]{x_1x_2...x_n}})$

Now using our lemma we have: $$(1+\frac{1}{x_1})(1+\frac{1}{x_2})...(1+\frac{1}{x_n})\geqslant \left(1+\frac{1}{\sqrt[n]{x_1x_2...x_n}}\right)^n$$

Then we only need to show that $\frac{1}{\sqrt[n]{x_1x_2...x_n}}\geqslant n\iff \frac{1}{n}\geqslant \sqrt[n]{x_1x_2...x_n}\iff\frac{x_1+x_2+...+x_n}{n}\geqslant\sqrt[n]{x_1x_2...x_n}$ because of our assumption $\sum_{k=1}^n x_k=1$

And the last inequality is true by the AM-GM.

Thus $(*)$ is proven.


We need prove: $(1+a)(1+b)(1+c) \geq (1+\sqrt[3]{abc})^3$

$\Leftrightarrow 1+abc+ab+bc+ca+a+b+c \geq 1+3\sqrt[3]{(abc)^2}+3\sqrt[3]{abc}+abc$

$\Leftrightarrow ab+bc+ca+a+b+c \geq 3\sqrt[3]{(abc)^2}+3\sqrt[3]{abc}$

Right by AM-GM. Apply we have:

$\left(1+\frac{1}{a} \right)\left(1+\frac{1}{b} \right)\left(1+\frac{1}{c} \right)=\dfrac{(1+a)(1+b)(1+c)}{abc} \geq \dfrac{(1+\sqrt[3]{abc})^3}{abc} \geq 64$

From $a+b+c=1 \Rightarrow abc\le \frac{1}{27}$

$$\Rightarrow \dfrac{(1+\sqrt[3]{abc})^3}{abc}=\bigg(\dfrac{1}{\sqrt[3]{abc}}+1\bigg)^3 \geq 64$$

When $a=b=c=\dfrac{1}{3}$


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