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I conjecture the following inequality:

$$\frac{\sqrt{\frac{\sum_{i=1}^n x_i^2}{n}}+\sqrt[n]{\prod_{i=1}^n x_i}}{2} \leq \frac{\sum_{i=1}^n x_i}{n}$$ for all real $x_1, x_2, \cdots, x_n > 0$

I am able to prove it when $n=2$.

Apply the power mean inequality between the $\frac12$power-mean and the arithmetic mean to $\frac{k^2+1}{2}$ and $k$ to obtain the following. Multiply by $x_1$ and let $x_2=kx_1$ to obtain the result.

$$\left(\frac{\sqrt{\frac{k^2+1}{2}}+\sqrt{k}}{2}\right)^2 \leq \frac{\frac{k^2+1}{2}+k}{2}$$

$$\left(\sqrt{\frac{k^2+1}{2}}+\sqrt{k}\right)^2 \leq k^2+1+2k$$

$$\sqrt{\frac{k^2+1}{2}}+\sqrt{k} \leq k+1$$

$$\sqrt{\frac{k^2x_1^2+x_1^2}{2}}+\sqrt{kx_1^2} \leq kx_1+x_1$$

$$\sqrt{\frac{x_2^2+x_1^2}{2}}+\sqrt{x_1x_2} \leq x_1+x_2$$

$$\frac{\sqrt{\frac{x_2^2+x_1^2}{2}}+\sqrt{x_1x_2}}{2} \leq \frac{x_1+x_2}{2}$$

I haven't made much progress with other $n$. I tried using the power mean inequality $\sqrt{\frac{\sum_{i=1}^n x_i^2}{n}} \leq \sqrt[n]{\frac{\sum_{i=1}^n x_i^n}{n}}$ for $n \geq 2$ to get the type of the radical the same on both sides, but that wouldn't work.

The conjecture is my own. Can someone prove or disprove it for general $n$?

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  • $\begingroup$ Your conjecture might hold true if you weight the LHS not $(\frac 1 2,\frac 1 2)$ but $(\frac 1 n,\frac {n-1} n)$. $\endgroup$ – A.S. Oct 19 '15 at 23:13
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Your inequality is false for $n\ge 3$ : A counterexample is $$(x_1,x_2,\cdots,x_{n-1},x_n)=(1,1,\cdots,1,2)$$

For $(x_1,x_2,\cdots,x_{n-1},x_n)=(1,1,\cdots,1,2)$, we have $$\frac{\sqrt{\frac{n-1+2^2}{n}}+2^{1/n}}{2}\le \frac{n-1+2}{n}.$$

This is equivalent to $$2+2x-\sqrt{1+3x}-2^x\ge 0$$ where $0\lt x=\frac 1n\le 1$.

Let $f(x)=2+2x-\sqrt{1+3x}-2^x.$ Then, $$f'(x)=2-\frac{3}{2\sqrt{1+3x}}-2^x\ln 2.$$

For $0\lt x\le \frac{1}{10}$, $$f'(x)=2-\frac{3}{2\sqrt{1+3x}}-2^x\ln 2\lt 2-\frac{3}{2\sqrt{\frac{3}{10}+1}}-\ln 2\lt 0.$$ With $f(0)=0$, we know that $f(x)\lt 0$ at least for $0\lt x\le \frac{1}{10}$.

Hence, we know that your inequality is false for $n\ge 10$. Checking each of $n=3,4,\cdots,9$ gives that your inequality is false for $n\ge 3$.

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For $n\geq 3$ the inequality can go either way.

Take $x_1=L\to \infty$ while others are equal to $1$ to get $$LHS>\frac L {2\sqrt n}+\frac 1 2 L^{1/n}>\frac L n+O(1)=RHS$$ which holds for $n\geq 4$. For $n=3$, consider a triplet $(8,1,1)$ to get

$$LHS=\frac 1 2\Big(\sqrt{\frac {64+2} 3}+2\Big)>\frac {10} 3=RHS$$

To get $LHS<RHS$ for any $n\geq 2$, let $x_1\to 0$ while others are equal to $1$: $$LHS\to\frac 1 2\sqrt\frac{n-1}n<\frac {n-1}n\leftarrow RHS$$

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