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

2018-03-12 20:24:03

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.

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

• 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)\geqslan

2018-03-12 22:43:42