# Mahler’s inequality

ចំពោះគ្រប់ចំនួនវិជ្ជមាន $a_1 ; a_2 ; ... ; a_n$ និង $b_1 ; b_2 ; ... ; b_n$ ដែល $2\leq n\in N$ គេបាន :

$\sqrt[n]{a_1a_2 ... a_n} + \sqrt[n]{b_1b_2 ... b_n}\leq \sqrt[n]{(a_1 + b_1)(a_2 + b_2) ... (a_n + b_n)}$

## សំរាយបញ្ជាក់

យើងមាន:

$\displaystyle \frac{a_1}{a_1 + b_1} + \frac{a_2}{a_2 + b_2} + ... + \frac{a_n}{a_n + b_n}\geq n\sqrt[n]{\frac{a_1a_2 ... a_n}{(a_1 + b_1)(a_2 + b_2) ... (a_n + b_n)}}$

$\displaystyle \frac{b_1}{a_1 + b_1} + \frac{b_2}{a_2 + b_2} + ... + \frac{b_n}{a_n + b_n}\geq n\sqrt[n]{\frac{b_1b_2 ... b_n}{(a_1 + b_1)(a_2 + b_2) ... (a_n + b_n)}}$

បូកអង្គនិងអង្គនៃវិសមភាពខាងលើយើងបាន

$\displaystyle 1 + 1 + ... + 1\geq \frac{n(\sqrt[n]{a_1a_2 ... a_n} + \sqrt[n]{b_1b_2 ... b_n})}{\sqrt[n]{(a_1 + b_1)(a_2 + b_2) ... (a_n + b_n)}}$

$\displaystyle n\geq \frac{n(\sqrt[n]{a_1a_2 ... a_n} + \sqrt[n]{b_1b_2 ... b_n})}{\sqrt[n]{(a_1 + b_1)(a_2 + b_2) ... (a_n + b_n)}}$

$\displaystyle \Longrightarrow \sqrt[n]{(a_1 + b_1)(a_2 + b_2) ... (a_n + b_n)}\geq \sqrt[n]{a_1a_2 ... a_n} + \sqrt[n]{b_1b_2 ... b_n}$

ត្រឡប់ទៅទំព័រ វិសមភាព Minkowski 1

បន្តទៅទំព័រ វិសមភាព Nesbitt