# Crux

គេអោយបណ្ដាចំនួនវិជ្ជមាន $a_1 ; a_2 ; ... ; a_n$ ដែលមានផលបូកស្មើ 1 ។

រកតំលៃតូចបំផុតនៃកន្សោមខាងក្រោម

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

## ចំលើយ

យើងមានពីររបៀបសំរាប់រកតំលៃតូចបំផុតនៃកន្សោមខាងលើ:

## 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 $a_1\leq a_2 \leq ... \leq a_n$

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

តាមវិសមភាព Chebyshev គេបាន

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

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

តាមវិសមភាព Bunhiacopski គេបាន

$(1.\sqrt{1 - a_1} + 1.\sqrt{1 - a_2} + ... + 1.\sqrt{1 - a_n})^2\leq n(n - a_1 - a_2 - ... - a_n)$

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

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

តាមវិសមភាព Cauchy – Schwarz គេបានវិសមភាពខាងក្រោម

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

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

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

ដូចនេះ $\displaystyle Min(S) = \sqrt{\frac{n}{n - 1}}$

## របៀបទី 2

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

$\displaystyle \Longrightarrow S = \biggl(\sqrt[3]{\frac{a_1}{\sqrt{1 - a_1}}}\biggl)^3 + \biggl(\sqrt[3]{\frac{a_2}{\sqrt{1 - a_2}}}\biggl)^3 + ... + \biggl(\sqrt[3]{\frac{a_n}{\sqrt{1 - a_n}}}\biggl)^3$

តាង $A = a_1(1 - a_1) + a_2(1 - a_2) + ... + a_n(1 - a_n)$

$\Longrightarrow A = \biggl(\sqrt[3]{a_1(1 - a_1)}\biggl)^3 + \biggl(\sqrt[3]{a_2(1 - a_2)}\biggl)^3 + ... + \biggl(\sqrt[3]{a_n(1 - a_n)}\biggl)^3$

តាមវិសមភាព Holder គេបាន

$S.S.A\geq (a_1 + a_2 + ... + a_n)^3 = 1$

$\Longrightarrow S^2.A\geq 1$

តែ $\displaystyle A = 1 - (a_1^2 + a_2^2 + ... + a_n^2)\leq 1 - \frac{(a_1 + a_2 + ... + a_n)^2}{n} = \frac{n - 1}{n}$

$\displaystyle \Longrightarrow \frac{1}{A}\geq \frac{n}{n - 1}$

ព្រោះ $\biggl((a_1 + a_2 + ... + a_n)^2\leq n(a_1^2 + a_2^2 + ... + a_n^2)\biggl)$

$\displaystyle \Longrightarrow S^2\geq \frac{1}{A}\geq \frac{n}{n - 1}$

$\displaystyle \Longrightarrow S\geq \sqrt{\frac{n}{n - 1}}$

ដូចនេះ $\displaystyle Min(S) = \sqrt{\frac{n}{n - 1}}$

## របៀបទី 3

តាង $\displaystyle f(x)=\frac{1}{\sqrt{1-x}}$

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

នាំអោយ $f(x)$ ជាអនុគមន៍ផតតាមវិសមភាព 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 $a_1+...+a_n=1$ គេបាន

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

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

$\displaystyle \Longrightarrow Min(S)=\sqrt{\frac{n}{n-1}}$

ដូចនេះវិសមភាពត្រូវបានស្រាយបញ្ជាក់។

## របៀបទី 4

តាង $\displaystyle f(x)=\frac{1}{\sqrt{1-x}}$ ដែល $x\in (0,1)$

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

គេបាន $f'(x)>0 \& f''(x) >0$

តាង $g(x) = xf(x)$

$g'(x)=f(x)+xf'(x)$

$g''(x)=f'(x)+f'(x)+xf''(x)=2f'(x)+xf''(x)>0$

គេបាន $g(x)$ ជាអនុគមន៍ផត។ តាមវិសមភាព Jensen គេបាន

$\displaystyle \frac{g(a_1)+g(a_2)+...+g(a_n)}{n}\geq g\biggl(\frac{a_1+a_2+...+a_n}{n}\biggl)$

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

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

ដោយ $a_1+a_2+...+a_n=1$ គេបាន

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

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

ដូចនេះ $\displaystyle Min=\sqrt{\frac{n}{n-1}}$

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