Proving that $\frac{\sqrt{1-x^2_{1}}+\sqrt{1-x^2_{2}}+\ldots+\sqrt{1-x^2_{17}}}{x_{1}+x_{2}+\ldots+x_{17}}\geq 4$

Question

If $x_{1},x_{2},x_{3},x_{4},\ldots,x_{17}\in \left[0,1\right]$ and $x^2_{1}+x^2_{2}+x^2_{3}+\ldots+x^2_{17} = 1$ then prove that $$\frac{\sqrt{1-x^2_{1}}+\sqrt{1-x^2_{2}}+\ldots+\sqrt{1-x^2_{17}}}{x_{1}+x_{2}+\ldots+x_{17}}\geq 4$$

$\bf{My\; Try:}$ Let $y^2_{i}=1-x_{i}^2,\forall i = 1,2,3,\ldots,17$

So $y^2_{1}+y^2_{2}+\ldots+y_{17}^2 = 16$

So here $y^2_{i}\leq y_{i},\forall i = 1,2 ,3,...,17$

So $$y^2_{1}+y^2_{2}+\ldots+y_{17}^2\leq y_{1}+y_{2}+y_{3}+\ldots+y_{17}$$

So $$y_{1}+y_{2}+y_{3}+\ldots+y_{17}\geq 16$$

Now how can I solve it after that? Help required.

Answer 1

Since $\sqrt{1-x_i^2}=\sqrt{\sum_{j=1, j\neq i}^{17} x_j^2}$, we have $$\frac{\sqrt{1-x^2_{1}}+\sqrt{1-x^2_{2}}+......+\sqrt{1-x^2_{17}}}{x_{1}+x_{2}+..........+x_{17}}=\frac{1}{\sum_{i=1}^{17} x_i}\left(\sqrt{\sum_{j=2}^{17} x_j^2}+\cdots+\sqrt{\sum_{j=1}^{16} x_j^2}\right).$$ By quadratic-arithmetic inequality, we have for $i\in\{1,\cdots 17\}$ $$\sqrt{\frac{\sum_{j=1, j\neq i}^{17} x_j^2}{16}}\geq \frac{\sum_{j=1, j\neq i}^{17} x_j}{16},$$ from where we get $$\sqrt{\sum_{j=1, j\neq i}^{17} x_j^2}\geq\frac{\sum_{j=1, j\neq i}^{17} x_j}{4}. $$ Finally, $$\begin{align}\frac{1}{\sum_{i=1}^{17} x_i}\left(\sqrt{\sum_{j=2}^{17} x_j^2}+\cdots+\sqrt{\sum_{j=1}^{16} x_j^2}\right)&\geq\frac{1}{\sum_{i=1}^{17} x_i}\left(\frac{\sum_{j=2}^{17} x_j}{4}+\cdots+\frac{\sum_{j=1}^{16} x_j}{4}\right)\\&=\frac{1}{\sum_{i=1}^{17} x_i}\left(\frac{16}{4}\sum_{j=1}^{17} x_j\right)\\[9pt]&=4.\end{align}$$

Answer 2

More generally, let $n$ be a positive integer, $x_1,x_2,\ldots,x_n$ nonnegative real numbers, and $S$ a subset of $\{1,2,\ldots,n\}$. For each $i=1,2,\ldots,n$, write $S+i$ for the set $$\big\{j\in\{1,2,\ldots,n\}\,\big|\,j\equiv s+i\pmod{n}\text{ for some }s\in S\big\}\,.$$ Then, for every real number $p\geq 1$, $$\sum_{i=1}^n\,\left(\sum_{j\in S+i}\,x_j^p\right)^{\frac{1}{p}}\geq\big(|S|\big)^{\frac{1}{p}}\,\sum_{i=1}^n\,x_i\,.$$ The equality holds if and only if any of the following conditions is met:
(1) $p=1$,
(2) $|S|=0$,
(3) $|S|=1$, or
(4) $x_1=x_2=\ldots=x_n$.
On the other hand, if $p\neq 0$ is a real number less than $1$, we have $$\sum_{i=1}^n\,\left(\sum_{j\in S+i}\,x_j^p\right)^{\frac{1}{p}}\leq\big(|S|\big)^{\frac{1}{p}}\,\sum_{i=1}^n\,x_i\,,$$ where the equality occurs if and only if any of the following conditions is true:
(1) $|S|=0$,
(2) $|S|=1$, or
(3) $x_1=x_2=\ldots=x_n$.
Both of these inequalities follow from the Power-Mean Inequality. We also have a version (via taking limit) for $p=0$ and $S\neq \emptyset$: $$\sum_{i=1}^n\,\left(\prod_{j\in S+i}\,x_j\right)^{\frac{1}{|S|}}\leq \sum_{i=1}^n\,x_i\,.$$ The equality happens if and only if $|S|=1$ or $x_1=x_2=\ldots=x_n$.

Answer 3

Let $x_i=\frac{a_i}{\sqrt{17}}$.

Hence, we need to prove that $\sum\limits_{i=1}^{17}\sqrt{17-a_i^2}\geq4\sum\limits_{i=1}^{17}a_i$, where $\sum\limits_{i=1}^{17}a_i^2=17$.

Indeed, we need to prove that $\sum\limits_{i=1}^{17}\left(\sqrt{17-a_i^2}-4a_i\right)\geq0$ or

$\sum\limits_{i=1}^{17}\frac{1-a_i^2}{4a_i+\sqrt{17-a_i^2}}\geq0$ or $\sum\limits_{i=1}^{17}\left(\frac{1-a_i^2}{4a_i+\sqrt{17-a_i^2}}+\frac{1}{8}\left(a_i^2-1\right)\right)\geq0$ or

$\sum\limits_{i=1}^{17}\frac{\left(a_i^2-1\right)\left(\sqrt{17-a_i^2}+4a_i-8\right)}{4a_i+\sqrt{17-a_i^2}}\geq0$ or $\sum\limits_{i=1}^{17}\frac{\left(a_i^2-1\right)\left(\sqrt{17-a_i^2}-4a_i+8a_i-8\right)}{4a_i+\sqrt{17-a_i^2}}\geq0$ or

$\sum\limits_{i=1}^{17}\frac{\left(a_i-1\right)^2\left(a_i+1\right)\left(8-\frac{17\left(a_i+1\right)}{4a_i+\sqrt{17-a_i^2}}\right)}{4a_i+\sqrt{17-a_i^2}}\geq0$ or $\sum\limits_{i=1}^{17}\frac{\left(a_i-1\right)^2\left(a_i+1\right)\left(15a_i+8\sqrt{17-a_i^2}-17\right)}{\left(4a_i+\sqrt{17-a_i^2}\right)^2}\geq0$,

for which it remains to prove that $8\sqrt{17-a_i^2}\geq17-15a_i$.

If $17-15a_i<0$ so the previous inequality is obvious.

But for $0\leq a_i\leq\frac{17}{15}$ it remains to prove that $64\left(17-a_i^2\right)\geq\left(17-15a_i\right)^2$ or

$47+30a_i-17a_i^2\geq0$, which is obvious for $0\leq a_i\leq\frac{17}{15}$. Done!