# Minkowski’s inequality

ចំពោះចំនួនវិជ្ជមាន $a_1 ; a_2 ; ... ; a_n$ និង $b_1 ; b_2 ; ... ; b_n ; \forall{n\in N^{*}}$

និងចំពោះ $p\geq 1$ គេបាន :

$\displaystyle \biggl(\sum_{i = 1}^{n}(a_i + b_i)^p\biggl)^{\frac{1}{p}}\leq \biggl(\sum_{i = 1}^{n}a_i^p\biggl)^{\frac{1}{p}}+\biggl(\sum_{i = 1}^{n}b_i^p\biggl)^{\frac{1}{p}}$

## សំរាយបញ្ជាក់

យើងមាន

$(a_i + b_i)^p = (a_i + b_i)^{p - 1}.(a_i + b_i) = a_i(a_i + b_i)^{p -1} + b_i(a_i + b_i)^{p - 1}$

តាមវិសមភាព Holder គេបាន

$\displaystyle \sum_{i = 1}^{n}a_i.(a_i + b_i)^{p - 1}\leq \biggl(\sum_{i = 1}^{n}a_i^p\biggl)^{\frac{1}{p}}.\biggl(\sum_{i = 1}^{n}(a_i + b_i)^{q(p - 1)}\biggl)^{\frac{1}{q}}$

$\displaystyle \sum_{i = 1}^{n}b_i.(a_i + b_i)^{p - 1}\leq \biggl(\sum_{i = 1}^{n}b_i^p\biggl)^{\frac{1}{p}}.\biggl(\sum_{i = 1}^{n}(a_i + b_i)^{q(p - 1)}\biggl)^{\frac{1}{q}}$

បូកអង្គនិងអង្គនៃវិសមភាពខាងលើគេបាន

$\displaystyle \sum_{i = 1}^{n}(a_i + b_i)^p\leq \biggl[\biggl(\sum_{i = 1}^{n}a_i^p\biggl)^{\frac{1}{p}} + \biggl(\sum_{i = 1}^{n}b_i^p\biggl)^{\frac{1}{p}}\biggl].\biggl(\sum_{i = 1}^{n}(a_i + b_i)^{q(p - 1)}\biggl)^{\frac{1}{q}}$

ដោយ $\displaystyle \frac{1}{p} + \frac{1}{q} = 1 \Longrightarrow q(p - 1) = p$ ជំនួសចូលវិសមភាពខាងលើគេបាន

$\displaystyle \sum_{i = 1}^{n}(a_i + b_i)^p\leq \biggl[\biggl(\sum_{i = 1}^{n}a_i^p\biggl)^{\frac{1}{p}} + \biggl(\sum_{i = 1}^{n}b_i^p\biggl)^{\frac{1}{p}}\biggl].\biggl(\sum_{i = 1}^{n}(a_i + b_i)^p\biggl)^{\frac{1}{q}}$

$\displaystyle \Longrightarrow \biggl(\sum_{i = 1}^{n}(a_i + b_i)^p\biggl)^{1 - \frac{1}{q}}\leq \biggl(\sum_{i = 1}^{n}a_i^p\biggl)^{\frac{1}{p}} + \biggl(\sum_{i = 1}^{n}b_i^p\biggl)^{\frac{1}{p}}$

ដោយ $\displaystyle \frac{1}{p} + \frac{1}{q} = 1 \Longrightarrow \frac{1}{p} = 1 - \frac{1}{q}$ ជំនួសចូលវិសមភាពខាងលើគេបាន

$\displaystyle \biggl(\sum_{i = 1}^{n}(a_i + b_i)^p\biggl)^{\frac{1}{p}}\leq \biggl(\sum_{i = 1}^{n}a_i^p\biggl)^{\frac{1}{p}}+\biggl(\sum_{i = 1}^{n}b_i^p\biggl)^{\frac{1}{p}}$

អ្នករៀបរៀង វ៉ាន់ ឃា

ត្រឡប់ទៅទំព័រ វិសមភាព Minkowski

បន្តទៅទំព័រ វិសមភាព Minkowski 2