Prove $(x+r_1) \cdots (x+r_n) \geq (x+(r_1 \cdots r_n)^{1/n})^{n}$.

Solution 1:

Here's a proof by induction: It holds trivially for $n=1$. For $n\ge2$ we have \begin{align*}(x+r_1)&\ldots(x+r_{n-1})(x+r_n)\ge(x+(r_1\ldots r_{n-1})^{1/(n-1)})^{n-1}(x+r_n)\\&=\Bigl(\bigl((x^{(n-1)/n})^{n/(n-1)}+((r_1\ldots r_{n-1})^{1/n})^{n/(n-1)}\bigr)^{(n-1)/n}\bigr((x^{1/n})^n+(r_n^{1/n})^n\bigr)^{1/n}\Bigr)^n\\&\ge\bigl(x+(r_1\ldots r_n)^{1/n}\bigr)^n.\end{align*} The last step is Hölder's inequality for $p=\frac{n-1}n$ and $q=\frac1n$.

Solution 2:

It's Holder inequality: $\left(x+(r_1r_2\cdots r_n)^{1/n}\right)^n = \left(x^{1/n}x^{1/n}\cdots x^{1/n}+r_1^{1/n}r_2^{1/n}\cdots r_n^{1/n}\right)^n \leq (x+r_1)(x+r_2)\cdots (x+r_n)$, and this is a special case of the following inequality:

$(a_1^n+b_1^n)(a_2^n+b_2^n)\cdots (a_n^n+b_n^n) \geq (a_1a_2\cdots a_n+b_1b_2\cdots b_n)^n$,

and when $n=2$ we get back our Cauchy-Buniakovski-Schwarz inequality.

Solution 3:

Expanding both sides as polynomials in $x$ and comparing the coefficients, the inequality just follows from Muirhead's inequality.

Solution 4:

Perhaps the simplest one is to use AM-GM only twice: $$n = \sum_{k=1}^n\dfrac{x}{x+r_k}+\sum_{k=1}^n\dfrac{r_k}{x+r_k}\geq n\dfrac{x}{(\prod_{k=1}^n(x+r_k))^{\frac{1}{n}}}+n\dfrac{(r_1r_2...r_n)^{\frac{1}{n}}}{(\prod_{k=1}^n(x+r_k))^{\frac{1}{n}}},$$ which then will yield the original inequality immmediately.