prove $\sum_{i=1}^{n}\sqrt{a_i}\ge (n-1)\sum_{i=0}^{n}\frac{1}{\sqrt{a_i}}$

Rearrangement works again!

Let $a_i=\frac{\sum\limits_{i=1}^nx_i-x_i}{x_i},$ where $i\in\{1,2,...,n-1\}$ and $x_1$, $x_2$,...,$x_n$ are positives.

Thus, $a_n=\frac{\sum\limits_{i=1}^nx_i-x_n}{x_n}$ and we need to prove that: $$\sum_{i=1}^n\sqrt{\frac{\sum\limits_{i=1}^nx_i-x_i}{x_i}}\geq(n-1)\sum_{i=1}^n\sqrt{\frac{x_i}{\sum\limits_{i=1}^nx_i-x_i}}$$ and since the last inequality is homogeneous, we can assume that $\sum\limits_{i=1}^nx_i=n$ and we need to prove that: $$\sum_{i=1}^n\sqrt{\frac{n-x_i}{x_i}}\geq(n-1)\sum_{i=1}^n\sqrt{\frac{x_i}{n-x_i}}$$ or $$\sum_{i=1}^n\left(\sqrt{\frac{n-x_i}{x_i}}-(n-1)\sqrt{\frac{x_i}{n-x_i}}\right)\geq0$$ or $$\sum_{i=1}^n\frac{1-x_i}{\sqrt{x_i(n-x_i)}}\geq0.$$ Now, let $x_1\leq x_2\leq...\leq x_n.$

Thus, for $i>j$ we have $$1-x_i\leq1-x_j$$ and $$\frac{1}{\sqrt{x_i(n-x_i)}}\leq\frac{1}{\sqrt{x_j(n-x_j)}}$$ because the last it's $$x_j(n-x_j)\geq x_i(n-x_i)$$ or $$(x_i-x_j)(n-x_i-x_j)\geq0,$$ which is obvious.

Thus, $(1-x_1,1-x_2,...,1-x_n)$ and $\left(\frac{1}{\sqrt{x_1(n-x_1)}},\frac{1}{\sqrt{x_2(n-x_2)}},...,\frac{1}{\sqrt{x_n(n-x_n)}}\right)$ have the same ordering and by Chebyshov we obtain: $$\sum_{i=1}^n\frac{1-x_i}{\sqrt{x_i(n-x_i)}}\geq\frac{1}{n}\sum_{i=1}^n(1-x_i)\sum_{i=1}^n\frac{1}{\sqrt{x_i(n-x_i)}}=0$$ and we are done!

We have that

$$\sum_{i=1}^{n}\sqrt{a_i}\ge ({n-1})\sum_{i=1}^{n}\frac{1}{\sqrt{a_i}} \iff \sum_{i=1}^{n}\left(\sqrt{a_i}+\frac{1}{\sqrt{a_i}}\right)\ge n\sum_{i=1}^{n}\frac{1}{\sqrt{a_i}}$$

and since

$$\sum_{i=1}^{n}\frac{1}{1+a_i}=\sum_{i=1}^{n}\frac{\frac{1}{\sqrt{a_i}}}{\frac{1}{\sqrt{a_i}}+\sqrt{a_i}}=1$$

by Chebyshev's Inequality we obtain

$$\sum_{i=1}^{n}\left(\sqrt{a_i}+\frac{1}{\sqrt{a_i}}\right)\cdot\sum_{i=1}^{n}\frac{\frac{1}{\sqrt{a_i}}}{\frac{1}{\sqrt{a_i}}+\sqrt{a_i}}\ge n\sum_{i=1}^{n}\frac{1}{\sqrt{a_i}}$$

indeed assuming wlog $a_i$ not decreasing by $x=\sqrt{a_i}$ we have that

• $f(x)=f\left(\frac1{x}\right)=x+\frac1x \implies f'(x)=1-\frac1{x^2}$ is not decreasing for $x\ge 1$
• $g(x)=\frac{\frac1x}{x+\frac1x}=\frac1{x^2+1}$ is not increasing

moreover we have that

$$\frac{1}{1+a_1}+\frac{1}{1+a_2}\le 1 \iff a_1a_2\ge 1 \iff \sqrt{a_1a_2}\ge 1\iff \sqrt{a_2}\ge\frac1{\sqrt{a_1}}$$

therefore if $a_1\le1$ we have that

$$f(\sqrt{a_1})= f\left(\frac1{\sqrt{a_1}}\right)\le f(\sqrt{a_2})$$

and the condition for the application of the inequality is preserved.

$\sum \limits_{i=i}^{n}\frac{1}{1+a_i}=1$ ...(i)

So $a_i \gt 0$ for all $i \in (1, n)$ and for $n \gt 1$.

We have to prove $\sum \limits_{i=1}^{n}\sqrt{a_i}\ge ({n-1})\sum \limits_{i=1}^{n}\frac{1}{\sqrt{a_i}}$

or prove, $\sum _\limits{i=1}^{n} \dfrac{1 + a_i}{\sqrt{a_i}} \ge n\sum \limits_{i=1}^{n}{\frac{1}{\sqrt a_i}}$

or prove, $(\sum _\limits{i=1}^{n} \dfrac{1 + a_i}{\sqrt{a_i}}) \, (\sum \limits_{i=1}^{n} \dfrac{1}{1+a_i}) \ge n\sum \limits_{i=1}^{n}\frac{1+a_i}{\sqrt a_i}{\frac{1}{1+a_i}}$ ...(ii)

WLOG, assume, $a_i \le a_{i+1}$ for $1 \le i \le (n-1)$.

Say, $f(a_i) = \dfrac {1+a_i}{\sqrt a_i} = \sqrt a_i + \dfrac{1}{\sqrt a_i}$. So, for all $a_i \gt 0$, $f(a_i) = f(\dfrac{1}{a_i})$

Say, $g(a_i) = \dfrac {1}{1+a_i}$. We can see $g(a_{i+1}) \le g(a_i)$.

We can see $g(a_i)$ is non-increasing. We now need to prove that $f(a_i)$ is non-decreasing. With that (ii) holds good as per Chebyshev's inequality.

It is easy to see that the function is non-decreasing for $a_i \gt 1$. It is also easy to see that $a_i$ can be $\lt 1$ only for one value of $i$, at max, due to given condition (i).

Say, $a_1 \lt 1$. We also know that $\dfrac {1}{1+a_2} \le 1 - \dfrac {1}{1+a_1} = \dfrac{1}{1+\dfrac{1}{a_1}}$.

So, $a_2 \ge \dfrac{1}{a_1}$ and $f(a_1) = f(\dfrac{1}{a_1}) \le f(a_2) \le ... \le f(a_n)$

With this, we prove that (ii) holds good as per Chebyshev's inequality.