Two Products: Constraint and Inequality
Problem
Proof 1
Since $\displaystyle\prod_{k=1}^na_k = 1,\,$
$\displaystyle\prod_{k=1}^n\left(1+\frac{1}{a_k}\right) = \prod_{k=1}^n(1+a_k)/\prod_{k=1}^na_k=\prod_{k=1}^n(1+a_k).$
Thus, if, say, $\displaystyle\prod_{k=1}^n(1+a_k)=A,\,$ then
$\displaystyle\begin{align} A^2 &= \prod_{k=1}^n\left(1+\frac{1}{a_k}\right)\prod_{k=1}^n(1+a_k)\\ &=\prod_{k=1}^n\left(1+\frac{1}{a_k}\right)(1+a_k)\\ &=\prod_{k=1}^n\left(1+1+a_k+\frac{1}{a_k}\right)\\ &\ge \prod_{k=1}^n(1+1+2) = 4^n\\ &= 2^{2n}, \end{align}$
implying that $A\ge 2^n,\,$ with equality only when $a_1=a_2=\ldots=a_n=1.$
Proof 2
By the AM-GM inequality,
$\displaystyle\prod_{k=1}^n(1+a_k)\ge\prod_{k=1}^n2\sqrt{a_k}=2^n\sqrt{\prod_{k=1}^na_k}=2^n.$
Proof 3
By Hölder's inequality,
$\displaystyle\prod_{k=1}^n(1+a_k)\ge\left(1+\sqrt[n]{\prod_{k=1}^na_k}\right)^n=2^n.$
Proof 4
That's a proof by induction.
For $n=1\,$ there is nothing to prove. Let's check the case of $n=2.\,$ We have $a_1a_2=1\,$ and need to prove that $(1+a_1)(1+a_2)\ge 4.\,$ But
$(1+a_1)(1+a_2)=1+a_1+a_2+a_1a_2=2+(a_1+a_2)\ge 2+2\sqrt{a_1a_2}=4.$
Assume the truth of the statement for some $n\ge 2\,$ and consider $n+1\,$ positive numbers such that $\displaystyle\prod_{k=1}^{n+1}a_k=1.\,$ One consequence of that condition is that there are bound to be two numbers, say, $a_{n}\le 1\,$ and $a_{n+1}\ge 1.\,$ By the inductive assumption, $\displaystyle\prod_{k=1}^{n-1}(1+a_k)(1+a_{n}a_{n+1})\ge 2^{n}.\,$ We thus have
$\displaystyle\begin{align} &0 \ge (1-a_{n-1})(1-a_n),\\ &a_{n-1}+a_n\ge 1 +a_{n-1}a_n,\\ &\frac{1}{2}(1+a_{n-1})(1+a_{n})\ge 1 +a_{n-1}a_n,\\ &\prod_{k=1}^{n-1}(1+a_k)\cdot \frac{1}{2}(1+a_{n-1})(1+a_{n})\ge\prod_{k=1}^{n-1}(1+a_k)(1 +a_{n-1}a_n)\ge 2^n,\\ &\prod_{k=1}^{n+1}(1+a_k)\ge 2^{n+1}. \end{align}$
Modification
Let $n\gt 0\,$ be an integer and $a_1, a_2, \ldots, a_n\,$ positive numbers such that $\displaystyle\prod_{k=1}^na_k = 1.\,$ Prove that $\displaystyle\prod_{k=1}^n(1+a_k+a_{k+1})\ge 3^n,$
where $a_{n+1}=a_1.$
Acknoweledgment
The problem came from an old Russian book (Ye. Vachovsky, A. Ryvkin, Problems in Elementary Mathematics of Increased Level of Difficulty, Nauka, 1971).
Solution 1 is from the book; Solutions 2 and 3 are by Leo Giugiuc.
Inequalities with the Product of Variables as a Constraint
- An Application of Schur's Inequality II $\left(\sum (x^4+y^3+z) \ge \sum \left(\frac{x^2+y^2}{z}\right)+3\right)$
- Problem 1 From the 2016 Pan-African Math Olympiad $\left(\displaystyle \sum_{cycl}\frac{1}{(x+1)^2+y^2+1}\le\frac{1}{2}\right)$
- A Cyclic But Not Symmetric Inequality in Four Variables $\left(\displaystyle 5(a+b+c+d)+\frac{26}{abc+bcd+cda+dab}\ge 26.5\right)$
- Problem 2, the 36th IMO (1995) $\left(\displaystyle \frac{1}{a^3(b+c)}+\frac{1}{b^3(c+a)}+\frac{1}{c^3(a+b)}\ge\frac{3}{2}\right)$
- An Inequality with Two Cyclic Sums $\left(\displaystyle \sum_{cycl}(a+\sqrt[3]{a}+\sqrt[3]{a^2})\ge 9\sum_{cycl}\frac{1}{1+\sqrt[3]{b^2}+\sqrt[3]{c}}\right)$
- The Roads We Take $(x(x-3(y+z))^2+(3x-(y+z))^2(y+z)\ge 27)$
- Long Huynh Huu's Inequality and Solution $\left(\displaystyle \sum_{i=1}^n\arctan x_i\le (n-1)\frac{\pi}{2}\right)$
- An Inequality with Integrals $\left(\Omega=\int_0^u\frac{u^4+7x^3-25x^2+37x+4}{x^4-3x^3+5x^2-3x+4}dx,\;\Omega(a)+\Omega(b)+\Omega(C)\ge3+5\ln\prod_{cycl}(1+a^2)\right)$
- An Inequality with Cycling Sums $\left(\displaystyle\sum_{cycl}(x^4+y^3+z)\ge \sum_{cycl}\frac{x^2+y^2}{z}+3\right)$
- Two Products: Constraint and Inequality $\left(\displaystyle\prod_{k=1}^n(1+a_k)\ge 2^n\right)$
- Leo Giugiuc's Cyclic Inequality with a Constraint $\left(\displaystyle a^3+b^3+c^3+\frac{16}{(a+b)(b+c)(c+a)}\ge 5\right)$
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