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Does $$a_k,b_k>0$$ imply that $$\left(\sum_{k=1}^n \frac{a_k}{n}\right)^2+\left(\sum_{k=1}^n \frac{b_k}{n}\right)^2\ge \prod_{k=1}^n(a_k^2+b_k^2)^{1/n}$$?

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The question I would now ask is whether $\left(\sum_{k=1}^n \frac{a_k}{n}\right)^2+\left(\sum_{k=1}^n \frac{b_k}{n}\right)^2\ge \frac{1}{2}\prod_{k=1}^n(a_k^2+b_k^2)^{1/n}$ –  robjohn Sep 10 '11 at 3:12

2 Answers 2

up vote 6 down vote accepted

Let $n = 2$. Take $a_1 = b_2 = 1$ and $a_2 = b_1 = \epsilon$.

Left hand side is $(1+\epsilon)^2 / 2$. Right hand side is $(1 + \epsilon^2)$.

So the inequality is false for any $\epsilon < 1$.

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No. Consider $n=2$ and $(a_k,b_k)=(1,0)\text{ and }(0,1)$.

The left side is the square of $$ \left|\frac{1}{n}\sum_{k=1}^n(a_k,b_k)\right|\tag{1} $$ and the right side is the square of $$ \left(\prod_{k=1}^n|(a_k,b_k)|\right)^\frac{1}{n}\tag{2} $$ If $(a_k,b_k)$ were allowed to be out of the first quadrant, it would be easy to cancel the sum in $(1)$ while the product in $(2)$ can only vanish if one of the $(a_k,b_k)=0$. However, enough cancellation occurs in $(1)$ if we place two unit vectors as far apart in the first quadrant as possible. Of course, the product in $(2)$ is $1$.

Actually, if $\{(a_k,b_k)\}$ is any set of at least $2$ distinct points on the unit circle in the first quadrant, $(1)$ and $(2)$ show that theses points disprove the posited inequality.

However, since $(a_k,b_k)\cdot(1,1)\ge|(a_k,b_k)|$, we get that $$ \left|\frac{1}{n}\sum_{k=1}^n(a_k,b_k)\right||(1,1)|\ge\frac{1}{n}\sum_{k=1}^n(a_k,b_k)\cdot(1,1)\ge\frac{1}{n}\sum_{k=1}^n|(a_k,b_k)|\ge\left(\prod_{k=1}^n|(a_k,b_k)|\right)^\frac{1}{n} $$ Therefore, we get that $$ \left(\sum_{k=1}^n \frac{a_k}{n}\right)^2+\left(\sum_{k=1}^n \frac{b_k}{n}\right)^2\ge \frac{1}{2}\prod_{k=1}^n(a_k^2+b_k^2)^{1/n} $$

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