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. 

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