If $x_i>a>0$ for $i=1,2\cdots n$ and $(x_1-a)(x_2-a)\cdots(x_n-a)=k^n$, $k>0$, prove by using the laws of inequality that $$x_1x_2 \cdots x_n\geq (a+k)^n$$.


If we expand $(x_1-a)(x_2-a)\cdots(x_n-a)=k^n$ in the LHS, we get $x_1x_2 \cdots x_n -a\sum x_1x_2\cdots x_{n-1} +a^2\sum x_1x_2\cdots x_{n-2} - \cdots +(-1)^na^n=k^n$. But it becomes cumbersome to go further. Please help me.

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    $\begingroup$ Start with $x_i-a=y_i$ $\endgroup$ – lab bhattacharjee Feb 14 '16 at 12:51
  • $\begingroup$ @labbhattacharjee Ok I understand. Thanks. $\endgroup$ – user1942348 Feb 14 '16 at 13:15
  • $\begingroup$ Note that a slightly stronger statement holds: if for each $i=1,2,\dots,n$ the real number $x_i$ is non-negative, $a$ is a real number between 0 and the minimum among the $x_i$, and $k$ is the geometric mean of the sequence $(x_i - a)$, then the geometric mean of the sequence $(x_i)$ is at least $a+k$. (It is not necessary to exclude the interval endpoints for $a$.) $\endgroup$ – András Salamon Mar 13 '18 at 10:38

From HUYGEN’S INEQUALITY, stating that for $x_i\geq0$ $$(1+x_1)(1+x_2)...(1+x_n)\geq\left(1+\left(x_1x_2...x_n\right)^{\frac{1}{n}}\right)^n \tag{1}$$ and (as it was suggested in comments) noting $x_i-a=y_i>0$ we have $$y_1y_2...y_n=k^n$$ as a result $$\color{red}{x_1x_2...x_n}=(y_1+a)(y_2+a)...(y_n+a)=\\ a^n\left(1+\frac{y_1}{a}\right)\left(1+\frac{y_2}{a}\right)...\left(1+\frac{y_n}{a}\right)\color{red}{\geq}\\ a^n\left(1+\left(\frac{y_1y_2...y_n}{a^n}\right)^{\frac{1}{n}}\right)^n= a^n\left(1+\left(\frac{k^n}{a^n}\right)^{\frac{1}{n}}\right)^n=\\ a^n\left(1+\frac{k}{a}\right)^n=\color{red}{(a+k)^n}$$

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    $\begingroup$ (+1) I tried to find out more about the origin of Huygens' inequality, but all the links, apart from the one you gave, were to trigonometric inequalities. Do you know anything more about where this came from? $\endgroup$ – John Bentin Mar 18 '18 at 10:11
  • $\begingroup$ One reference is artofproblemsolving.com/community/…. I personally learned it long time ago, ~1994 from a IMO reference material. $\endgroup$ – rtybase Mar 18 '18 at 21:06

Using convenient notation, we will prove a theorem from which the result of the question may be easily derived. Given any positive real numbers $ x_1, x_2,...,$ we will write their initial geometric means as $$g_n:=(x_1\cdots x_n)^{1/n}\quad(n=1,2,...).$$

Theorem.$\quad$Given any $a\geqslant0$, and for each $n$, the following proposition (which we will denote by $P_n$) holds:$$(x_1+a)\cdots(x_n+a)\geqslant (g_n+a)^n\quad\text{for all}\quad x_1,x_2,...>0.$$Proof.$\quad$We proceed by Cauchy induction. That is, we establish (1) $P_1\,;\;$ (2) $P_k\Rightarrow P_{2k}$ for any $k=1,2,...;$ and (3) $P_{k+1}\Rightarrow P_k$ for any $k=1,2,...$.

(1)$\quad$Clearly the inequality (actually equality) holds for $n=1$ since $g_1=x_1$ in this case.

(2)$\quad$To prove this, let us suppose that $P_n$ has been established for the case $n=k$:$$(x_1+a)\cdots(x_k+a)\geqslant(g_k+a)^k.$$A corresponding result holds also for $x_{k+1},...,x_{2k}$ , which we write as$$(x_{k+1}+a)\cdots(x_{2k}+a)\geqslant\left[\left(\frac{x_1\cdots x_{2k}}{x_1\cdots x_k}\right)^{1/k}+a\right]^k=\left(\frac{g_{2k}^2}{g_k}+a\right)^k.$$Composing the above two inequalities gives$$(x_1+a)\cdots(x_{2k}+a)=(x_1+a)\cdots(x_k+a)\cdot(x_{k+1}+a)\cdots(x_{2k}+a)$$$$\geqslant(g_k+a)^k(g_{2k}^2/g_k+a)^k$$$$\qquad=[g_{2k}^2+a(g_k+g_{2k}^2/g_k)+a^2]^k$$$$\geqslant(g_{2k}+a)^{2k},$$where the last inequality follows by observing that $$ g_k+g_{2k}^2/ g_k=(\surd g_k-g_{2k}/\surd g_k)^2+2g_{2k}\geqslant 2g_{2k}$$(or by applying AM–GM to $g_k$ and $g_{2k}^2/g_k$). Thus we have established $P_k\Rightarrow P_{2k}$ .

(3)$\quad$It remains to show $P_{k+1}\Rightarrow P_k$ . Suppose, then, that $P_{k+1}$ holds for some $k\in\{1,2,...\}$: $$(x_1+a)\cdots(x_k+a)(x_{k+1}+a)\geqslant(g_{k+1}+a)^{k+1}$$for all $x_1,x_2,...>0.$ Then, in particular, it holds in the case when $x_{k+1}=g_k$. In this case,$$g_{k+1}=(g_k^kx_{k+1})^{1/(k+1)}=g_k.$$ Therefore, dividing by the factor $g_k+a\,$ (or $x_{k+1}+a$) gives $(x_1+a)\cdots(x_k+a)\geqslant(g_k+a)^k,$ which is the statement of $P_k$.$\quad\square$

The result of the question can be obtained via the replacement of $x_i+a$ by $x_i\;\,(i=1,...,n)$ and of $g_n$ by $k$ in the statement of $P_n$.


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