# (China MO 1996)

គេអោយ $a_1 ; a_2 ; ... ; a_n$ ជាចំនួនវិជ្ជមានដែលមានផលបូកស្មើ 1 ។ ស្រាយថា:

$\displaystyle \frac{a_1}{\sqrt{1 - a_1}} + \frac{a_2}{\sqrt{1 - a_2}} + ... + \frac{a_n}{\sqrt{1 - a_n}}\geq \frac{\sqrt{a_1} + \sqrt{a_2} + ... + \sqrt{a_n}}{\sqrt{n - 1}}$

## ចំលើយ

របៀបទី 1

តាង $\displaystyle S = \frac{a_1}{\sqrt{1 - a_1}} + \frac{a_2}{\sqrt{1 - a_2}} + ... + \frac{a_n}{\sqrt{1 - a_n}}$

យើងឧបមាថា $a_1\leq a_2\leq ... \leq a_n$

$\displaystyle \frac{1}{\sqrt{1 - a_1}}\leq \frac{1}{\sqrt{1 - a_2}}\leq ... \leq \frac{1}{\sqrt{1 - a_n}}$

តាមវិសមភាព chebyshev គេបាន

$\displaystyle S\geq \frac{1}{n}.(a_1 + a_2 + ... + a_n)(\frac{1}{\sqrt{1 - a_1}} + \frac{1}{\sqrt{1 - a_2}} + ... + \frac{1}{\sqrt{1 - a_n}})$

$\displaystyle S\geq \frac{1}{n}.(\frac{1}{\sqrt{1 - a_1}} + \frac{1}{\sqrt{1 - a_2}} + ... + \frac{1}{\sqrt{1 - a_n}})$

តែតាមវិសមភាព Bunhiacopski គេបាន

$n = n(a_1 + a_2 + ... + a_n)\geq (\sqrt{a_1} + \sqrt{a_2} + ... + \sqrt{a_n})^2$

$\Longrightarrow \sqrt{a_1} + \sqrt{a_2} + ... + \sqrt{a_n}\leq \sqrt{n}$

$\sqrt{1 - a_1} + \sqrt{1 - a_2} + ... + \sqrt{1 - a_n}\leq \sqrt{n(n - a_1 - a_2 - ... - a_n)}$

$\Longrightarrow \sqrt{1 - a_1} + \sqrt{1 - a_2} + ... + \sqrt{1 - a_n}\leq \sqrt{n(n - 1)}$

យើងទាញបាន

$\displaystyle \frac{1}{\sqrt{1 - a_1} + \sqrt{1 - a_2} + ... + \sqrt{1 - a_n}}\geq \frac{1}{\sqrt{n(n - 1)}}$

តាមវិសមភាព Cauchy – Schwarz គេបានវិសមភាព

$\displaystyle \frac{1}{\sqrt{1 - a_1}} + \frac{1}{\sqrt{1 - a_2}} + ... + \frac{1}{\sqrt{1 - a_n}}\geq \frac{(1 + 1 + ... + 1)^2}{\sqrt{1 - a_1} + \sqrt{1 - a_2} + ... + \sqrt{1 - a_n}}$

$\displaystyle \Longrightarrow \frac{1}{\sqrt{1 - a-1}} + \frac{1}{\sqrt{1 - a_2}} + ... + \frac{1}{\sqrt{1 - a_n}}\geq \frac{n^2}{\sqrt{n(n - 1)}}$

$\displaystyle \Longrightarrow S\geq \frac{1}{n}.\frac{n^2}{\sqrt{n(n - 1)}} = \frac{\sqrt{n}}{\sqrt{n - 1}}$

តែ $\sqrt{a_1} + \sqrt{a_2} + ... + \sqrt{a_n}\leq \sqrt{n}$ សំរាយខាងលើ។

$\displaystyle \Longrightarrow S\geq \frac{\sqrt{a_1} + \sqrt{a_2} + ... + \sqrt{a_n}}{\sqrt{n - 1}}$

ដូចនេះ  $\displaystyle \frac{a_1}{\sqrt{1 - a_1}} + \frac{a_2}{\sqrt{1 - a_2}} + ... + \frac{a_n}{\sqrt{1 - a_n}}\geq \frac{\sqrt{a_1} + \sqrt{a_2} + ... + \sqrt{a_n}}{\sqrt{n - 1}}$

របៀបទី 2

តាង $\displaystyle f(x)=\frac{1}{\sqrt{1-x}}, x<1$ $\displaystyle f'(x)=\frac{1}{2}(1-x)^{-\frac{3}{2}}\Longrightarrow f''(x)=\frac{3}{4}(1-x)^{-\frac{5}{2}}>0$

តាមវិសមភាព V-K ចំពោះ $k=1$ គេបាន

$\displaystyle a_1f(a_1)+...+a_nf(a_n)\geq (a_1+...+a_n)f\biggl(\frac{a_1+...+a_n}{n}\biggl)$

ដោយ $a_1+...+a_n=1$

$\displaystyle \Longrightarrow a_1f(a_1)+...+a_nf(a_n)\geq f\biggl(\frac{1}{n}\biggl)$

$\displaystyle \frac{a_1}{\sqrt{1-a_1}}+...+\frac{a_n}{\sqrt{1-a_n}}\geq \frac{1}{\sqrt{1-\frac{1}{n}}}=\frac{\sqrt{n}}{\sqrt{n-1}}$

តែតាមវិសមភាព Bunhiacopski គេបានវិសមភាព

$\sqrt{a_1}+...+\sqrt{a_n}\leq \sqrt{n}(a_1+...+a_n)=\sqrt{n}$

$\Longrightarrow \sqrt{n}\geq \sqrt{a_1}+...+\sqrt{a_n}$

តាមវិសមភាពខាងលើគេបាន
$\displaystyle \frac{a_1}{\sqrt{1-a_1}}+...+\frac{a_n}{\sqrt{1-a_n}}\geq \frac{\sqrt{a_1}+...+\sqrt{a_n}}{\sqrt{n-1}}$

ដូចនេះវិសមភាពត្រូវបានស្រាយបញ្ជាក់។