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

I can show that for $x > 0$ and $r_{i} > 0$ we have $$\left(\, x + r_{1}\,\right)\ldots\left(\, x + r_{n}\,\right)\ \geq\ \left[\, x + \left(\, r_{1}\ldots r_{n}\,\right)^{1/n}\,\right]^{n}.$$ However, I can't do this using straight up induction, strong or weak. Can someone do this?

You can show the inequality holds for $n$ if it holds for $n+1$ using $r_{n+1} = (r_1 \cdots r_n)^{1/n}$, so going down isn't a problem, but going up has eluded me. Additionally, you can show that if it holds for $n=a$ and $n=b$ then it holds for $n=ab$. Using powers of two and the above is sort of a way to prove the above by induction but it certainly isn't "normal".

You can also prove the above using Lagrange multipliers (which isn't surprising). The last proof I know is where you compare coefficients of $x^k$ and use the AM-GM inequality.

• I know three proofs already but if someone can show me one I haven't seen I'd be glad to see it! – abnry Dec 23 '14 at 1:00
• And what proofs do you know? Do you want people wasting their time on something you already know? – David Peterson Dec 23 '14 at 1:13
• @DavidPeterson, others don't know the proofs, possibly. These sorts of inequalities seems to have myriads of proof. Anyways, what I care most about it is proving this with induction. – abnry Dec 23 '14 at 1:15

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 $q=\frac1n$.

• Alright, I get this! Very nice! – abnry Dec 23 '14 at 1:46

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.

• I didn't think of generalized holder! – abnry Dec 23 '14 at 1:16

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

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.