If $(x_1-a)(x_2-a)\cdots(x_n-a)=k^n$ prove by using the laws of inequality that $x_1x_2 \cdots x_n\geq (a+k)^n$ 
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 $$x_1x_2 \cdots x_n\geq (a+k)^n$$.

Attempt:
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. 
 A: 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 $P_n$.
A: 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
$$\color{red}{x_1x_2...x_n}=(y_1+a)(y_2+a)...(y_n+a)=\\
a^n\left(1+\frac{y_1}{a}\right)\left(1+\frac{y_2}{a}\right)...\left(1+\frac{y_n}{a}\right)\color{red}{\geq}\\
a^n\left(1+\left(\frac{y_1y_2...y_n}{a^n}\right)^{\frac{1}{n}}\right)^n=
a^n\left(1+\left(\frac{k^n}{a^n}\right)^{\frac{1}{n}}\right)^n=\\
a^n\left(1+\frac{k}{a}\right)^n=\color{red}{(a+k)^n}$$
