How to prove this inequality $\sum_{r=1}^{n}a_{r}\sqrt{\frac{n-1}{1-a_{r}}}\ge\sum_{r=1}^{n}\sqrt{a_{r}}$

Question

Let $({a_{r}})_{r=1}^{n}$ be a sequence of $n$ positive real numbers that sum to 1. Prove that for all $n>1$ :\begin{align} \sum_{r=1}^{n}a_{r}\sqrt{\frac{n-1}{1-a_{r}}}&\ge\sum_{r=1}^{n}\sqrt{a_{r}}. \end{align}

$(x_1, x_2, \ldots, x_n)$, and $(y_1, y_2, \ldots, y_n)$, $$\left(\sum_{i=1}^{n} x_iy_i\right)^2 \leq \left(\sum_{i=1}^{n} x_i^2\right)\left(\sum_{i=1}^{n} y_i^2\right).$$

$$\begin{align*} x_i &= a_i\sqrt{\frac{n-1}{1-a_i}}, \ y_i &= \sqrt{a_i}. \end{align*}$$

Using cauchy-Schwarz inequality

$$\left(\sum_{i=1}^{n} a_i\sqrt{\frac{n-1}{1-a_i}}\cdot \sqrt{a_i}\right)^2 \leq \left(\sum_{i=1}^{n} \left(a_i\sqrt{\frac{n-1}{1-a_i}}\right)^2\right)\left(\sum_{i=1}^{n} \left(\sqrt{a_i}\right)^2\right).$$

$$\left(\sum_{i=1}^{n} a_i\sqrt{\frac{n-1}{1-a_i}}\cdot \sqrt{a_i}\right)^2 \leq \left(\sum_{i=1}^{n} \frac{n-1}{1-a_i}\cdot a_i\right)\left(\sum_{i=1}^{n} a_i\right).$$

$$\left(\sum_{i=1}^{n} a_i\sqrt{\frac{n-1}{1-a_i}}\cdot \sqrt{a_i}\right)^2 \leq \left(\sum_{i=1}^{n} \frac{n-1}{1-a_i}\cdot a_i\right)$$

$$\sum_{i=1}^{n} \frac{n-1}{1-a_i}\cdot a_i = (n-1)\sum_{i=1}^{n} \frac{a_i}{1-a_i}.$$

Sir Michael Rozenberg proved this with Jensen's inequality but can someone help me to prove this inequality with holder's inequality

Answer 1

Another way.

From my previous post we need to prove that: $$\sum_{r=1}^na_r\sqrt{\frac{n-1}{n-a_r}}\geq\sum_{r=1}\sqrt{a_r},$$ where $\sum\limits_{r=1}^na_r=n$ and $a_r>0$.

Now, by C-S $$n=\sqrt{n^2}=\sqrt{\sum_{r=1}^n1\sum_{r=1}^n}a_r\geq\sum_{r=1}^n\sqrt{a_r}$$ and by Holder we obtain: $$\sum_{r=1}^na_r\sqrt{\frac{n-1}{n-a_r}}=\sqrt{\frac{(n-1)\left(\sum\limits_{r=1}^n\frac{a_r}{\sqrt{n-a_r}}\right)^2\sum\limits_{r=1}^na_r(n-a_r)}{\sum\limits_{r=1}^na_r(n-r)}}\geq$$ $$\geq\sqrt{\frac{(n-1)\left(\sum\limits_{r=1}^na_r\right)^3}{\sum\limits_{r=1}^na_r(n-a_r)}}=\sqrt{\frac{(n-1)n^3}{\sum\limits_{r=1}^n(na_r-a_r^2-n+1)+n(n-1)}}=$$ $$=\sqrt{\frac{(n-1)n^3}{\sum\limits_{r=1}^n(a_r-1)(n-1-a_r)+n(n-1)}}=$$ $$=\sqrt{\frac{(n-1)n^3}{\sum\limits_{r=1}^n(a_r-1)(n-1-a_r-n+2)+n(n-1)}}=$$ $$=\sqrt{\frac{(n-1)n^3}{-\sum\limits_{r=1}^n(a_r-1)^2+n(n-1)}}\geq\sqrt{\frac{(n-1)n^3}{n(n-1)}}=n\geq\sum_{r=1}^n\sqrt{a_r}$$ and we are done.

Answer 2

For $\sum\limits_{r=1}^na_r=n$ we need to prove that: $$\sum_{r=1}^n\frac{a_r}{n}\sqrt{\frac{n-1}{1-\frac{a_r}{n}}}\geq\sum_{r=1}^n\sqrt{\frac{a_r}{n}}$$ or $$\sum_{r=1}^na_r\sqrt{\frac{n-1}{n-a_r}}\geq\sum_{r=1}\sqrt{a_r}.$$ Now, let $f(x)=x\sqrt{\frac{n-1}{n-x}}-\sqrt{x}.$

Thus, $$f''(x)=\left(\frac{\sqrt{n-1}(x-n+n)}{\sqrt{n-x}}-\sqrt{x}\right)''=$$ $$=\left(\sqrt{n-1}\left(-\sqrt{n-x}+\frac{n}{\sqrt{n-x}}\right)-\sqrt{x}\right)''=$$ $$=\sqrt{n-1}\left(\frac{1}{4\sqrt{(n-x)^3}}+\frac{3n}{4\sqrt{(n-x)^5}}\right)+\frac{3}{4\sqrt{x^3}}>0$$ and your inequality follows from Jensen.