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

Question

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

My try: i tried substituiting $y_i=\frac{1}{1+a_i}$ thus $\sum y_i=1$

also rearranging inequality we have to prove $$\sum_{i=1}^{n}(\sqrt{a_i}+\frac{1}{\sqrt{a_i}})\ge n\sum_{i=1}^{n}\frac{1}{\sqrt{a_i}}$$

putting value in terms of $y_i$ $$\sum_{i=1}^{n}\frac{1}{\sqrt{y_i(1-y_i)}}\ge n\sum_{i=1}^{n}\frac{\sqrt{y_i}}{\sqrt{1-y_i}}$$.i am stuck now i dont know which inequality to use.i tried using A-M$\ge$ H-M inequality.

source ' inequalities A mathematical Olympiad approach'

Answer 1

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

Answer 2

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!

Answer 3

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.

Answer 4

For this solution, I will only use this inequality, AM-GM Inequality and C-S. We need to prove that: $$\sum_{i=1}^{n}\sqrt{a_i}\ge ({n-1})\sum_{i=1}^{n}\frac{1}{\sqrt{a_i}}$$ if $$\sum_{i=1}^{n}\frac{1}{1+a_i}=1$$

Case I: Assume $a_i\geq1$ for all $i$ till $n$

$$1=\sum_{i=1}^{n}\frac{1}{1+a_i}\geq {n^2\over n+\sum_{i=0}^n{a_i}}\Rightarrow \sum_{i=0}^n{a_i}\geq n(n-1)\Rightarrow n+\sum_{i=0}^n{a_i}\geq n(n-1)+n$$

$$ \Rightarrow 2\sum_{i=0}^n{\sqrt{a_i}}\geq n(n-1)+n $$

As $\sum_{i=0}^n{\sqrt{a_i}}\geq n$,

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

Now it's enough to prove that

$$ n(n-1)\geq (n-1)\sum_{i=1}^{n}\frac{1}{\sqrt{a_i}} \Rightarrow n \geq \sum_{i=1}^{n}\frac{1}{\sqrt{a_i}}$$ which is obvious for this case.

Case II:
Assume that some $a_i \ge 1$ and others $\le 1$.
Note that, as mentioned in @MathLover's solution, only one $a_i$ can be $\leq 1$. Let $a_1$ be less than $1$ and rename others $a_{2},...,a_n$ where $1\leq k\leq n$.
Just to reduce this case hence the solution, the cases that arise in this case are left, as resolving them is very easy, in the sense that they are similar to resolving case I as before. (and the other is very easy) Just assume $$\frac{1}{1+a_1}=p,0\leq p\leq 1$$ for one case in this case, and $$\sum_{i=2}^{n}\frac{1}{1+a_i}=1-p,0\leq 1-p\leq 1$$ for the other. Just prove the original inequality for these two cases, and sum up the hypotheses and the inequalities to get what was desired. Done!