លំហាត់លីមីត

គណនាលីមីត \displaystyle \lim_{x\to 0}\frac{1-cosaxcosbx}{x^2}

\displaystyle \lim_{x\to 0}\frac{1-cosaxcosbx}{x^2}=\lim_{x\to 0}\frac{1-cosax+cosax(1-cosbx)}{x^2}

\displaystyle =\lim_{x\to 0}\frac{1-cosax}{x^2}+\lim_{x\to 0}\frac{1-cosbx}{x^2}cosax=\frac{a^2+b^2}{2}

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វិសមភាពគួរយល់ដឹង

ចំណុចទី 1

ចំពោះចំនួនវិជ្ជមាន a_1, a_2, ..., a_n និង b_1, b_2, ..., b_n ដែលផ្ទៀងផ្ទាត់

a_1\leq a_2\leq ...\leq a_n និង b_1\geq b_2\geq ...\geq b_n នោះគេបានវិសមភាពខាងក្រោម

\displaystyle \frac{a_1}{b_1}+\frac{a_2}{b_2}+...+\frac{a_n}{b_n}\geq \frac{n(a_1+a_2+...+a_n)}{b_1+b_2+...+b_n}

ចំនុចទី 2

ចំពោះចំនួនវិជ្ជមាន x_1, x_2, ... , x_n គេបាន

\displaystyle x_1^p+x_2^p+...+x_n^p\geq n\biggl(\frac{x_1+x_2+...+x_n}{n}\biggl)^p ; p\geq 2

លំហាត់សំរាប់សិស្សពូកែ

គេអោយ a, b, c, d ជាចំនួនវិជ្ជមានដែល a+b+c+d=1 ។ គណនាតំលៃតូចបំផុតនៃកន្សោម:

\displaystyle A=\frac{a^2\sqrt{1+a^2}}{1-a}+\frac{b^2\sqrt{1+b^2}}{1-b}+\frac{c^2\sqrt{1+c^2}}{1-c}+\frac{d^2\sqrt{1+d^2}}{1-d}

សំរាយបញ្ជាក់

ដោយ a+b+c+d=1\Longrightarrow a+b+c=1-d, b+c+d=1-a, c+d+a=1-b, d+a+b=1-c

ជំនួសចូលវិសមភាពខាងលើយើងបានៈ

\displaystyle A=\frac{a^2\sqrt{1+a^2}}{b+c+d}+\frac{b^2\sqrt{1+b^2}}{c+d+a}+\frac{c^2\sqrt{1+c^2}}{d+a+b}+\frac{d^2\sqrt{1+d^2}}{a+b+c}

យើងឧបមាថា a\leq b\leq c\leq d\Longrightarrow a+b+c\leq d+a+b\leq c+d+a\leq b+c+d

នោះយើងបាន: \displaystyle \frac{a^2}{b+c+d}\leq \frac{b^2}{c+d+a}\leq \frac{c^2}{d+b+a}\leq \frac{d^2}{a+b+c}

តាង \displaystyle f(x)=\sqrt{1+x^2}= (1+x^2)^{\frac{1}{2}} , x\geq 0

\displaystyle f'(x)=\frac{1}{2}.2x.(1+x^2)^{-\frac{1}{2}}=x(1+x^2)^{-\frac{1}{2}}>0

\displaystyle f''(x)=(1+x^2)^{-\frac{1}{2}}-\frac{1}{2}.2x^2(1+x^2)^{-\frac{3}{2}}=(1+x^2)^{-\frac{3}{2}}>0

តាមវិសមភាពទាំងប្រាំបី ចំពោះ a_1\leq a_2\leq a_3\leq a_4 ; b_1\leq b_2\leq b_3\leq b_4 និង f'(x)>0, f''(x)>0 គេបាន៖

\displaystyle b_1f(a_1)+b_2f(a_2)+b_3f(a_3)+b_4f(a_4)\geq (b_1+...+b_4)f\biggl(\frac{a_1+...+a_4}{4}\biggl)

ទ្រឹស្ដីបទ

ចំពោះចំនួនវិជ្ជមាន a_1, a_2, ...a_{m-1}, a_m, a_{m+1},..., a_{n-1}, a_n គេបាន

\displaystyle \frac{a_1}{a_2}+\frac{a_2}{a_3}+...+\frac{a_{n-1}}{a_n}+\frac{a_n}{a_1}\geq n

\displaystyle \frac{a_1}{a_2+a_3}+\frac{a_2}{a_3+a_4}+...+\frac{a_{n-1}}{a_n+a_1}+\frac{a_n}{a_1+a_2}\geq \frac{n}{2}

\displaystyle \frac{a_1}{a_2+a_3+a_4}+\frac{a_2}{a_3+a_4+a_5}+...+\frac{a_{n-1}}{a_n+a_1+a_2}+\frac{a_n}{a_1+a_2+a_3}\geq \frac{n}{3}

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យើងអាចសរសេរការបូកវិសមភាពខាងលើជាទំរង់ទូទៅដូចខាងក្រោម

ចំពោះ m\leq n-1 គេបាន

\displaystyle \frac{a_1}{a_2+a_3+...+a_{m+1}}+\frac{a_2}{a_3+a_4+...+a_{m+2}}+...+\frac{a_{n-1}}{a_n+a_1+...+a_{m-1}}+\frac{a_n}{a_1+a_2+...+a_m}

\displaystyle \geq \frac{n}{m}

នេះជាវិសមភាពមួយដែលខ្ញុំទើបតែនឹងបានធ្វើការស្រាយបញ្ជាក់រួច។ ហើយចំពោះសំរាយបញ្ជាក់នោះ

ខ្ញុំមិនទាន់បានផុសទេ គឺទុកអោយអ្នកចូលចិត្តគណិតវិទ្យារៀនស្រាយលេងសិន 😀

និពន្ធដោយ វ៉ាន់ ឃា

លំហាត់មានលក្ខខណ្ឌ

ស្រាយបញ្ជាក់ថាចំពោះចំនួនវិជ្ជមាន a\leq b\leq c ដែល a^2+b^2+c^2=2 គេបាន:

\displaystyle \frac{1-a^2}{c(a+b)}+\frac{1-b^2}{a(b+c)}+\frac{1-c^2}{b(c+a)}\geq \frac{3}{4}

សំរាយបញ្ជាក់

វិសមភាពខាងលើសមមួលនឹង

\displaystyle \frac{2-2a^2}{c(a+b)}+\frac{2-2b^2}{a(b+c)}+\frac{2-2c^2}{b(c+a)}\geq \frac{3}{2}

ដោយ a^2+b^2+c^2=2 នោះវិសមភាពខាងលើទៅជា

\displaystyle \frac{a^2+b^2+c^2-2a^2}{c(a+b)}+\frac{a^2+b^2+c^2-2b^2}{a(b+c)}+\frac{a^2+b^2+c^2-2c^2}{b(c+a)}\geq \frac{3}{2}

\displaystyle \frac{b^2-a^2+c^2}{c(a+b)}+\frac{a^2-b^2+c^2}{a(b+c)}+\frac{a^2+b^2-c^2}{b(c+a)}\geq \frac{3}{2}

\displaystyle \frac{(b-a)(a+b)+c^2}{c(a+b)}+\frac{(c-b)(b+c)+a^2}{a(b+c)}+\frac{(a-c)(c+a)+b^2}{b(c+a)}\geq \frac{3}{2}

\displaystyle \frac{b-a}{a}+\frac{c}{a+b}+\frac{c-b}{c}+\frac{a}{b+c}+\frac{a-c}{b}+\frac{b}{c+a}\geq \frac{3}{2}

\displaystyle \frac{c-b}{a}-\frac{c-a}{b}+\frac{b-a}{c}+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{3}{2} ; (I)

តាង \displaystyle f(x)=\frac{1}{x} , x>0

\displaystyle f'(x)=-\frac{1}{x^2}

\displaystyle f''(x)=\frac{2}{x^3}>0

តាមវិសមភាព V – K ចំពោះ a\leq b\leq c និង f''(x)>0 គេបាន:

(c-b)f(a)-(c-a)f(b)+(b-a)f(c)\geq 0

\displaystyle \Longrightarrow \frac{c-b}{a}-\frac{c-a}{b}+\frac{b-a}{c}\geq 0

\displaystyle \Longrightarrow \frac{c-b}{a}-\frac{c-a}{b}+\frac{b-a}{c}+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} ; (II)

តាមវិសមភាព Nesbitt គេមាន \displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{3}{2}

តាម (II) នាំអោយគេបាន:

\displaystyle \frac{c-b}{a}-\frac{c-a}{b}+\frac{b-a}{c}+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{3}{2}  ; (III)

តាម (III) យើងទទួលបាន (I)

ដូចនេះវិសមភាពត្រូវបានស្រាយបញ្ជាក់។

លំហាត់ទើបច្នៃ

គេអោយបីចំនួនវិជ្ជមាន a\leq b\leq c ។ ស្រាយបញ្ជាក់ថា

\displaystyle 3\biggl(\frac{a^3}{b^3}+\frac{b^3}{c^3}+\frac{c^3}{a^3}\biggl)^2\geq \biggl(\frac{b^2}{a^2}+\frac{c^2}{b^2}+\frac{a^2}{c^2}\biggl)^3

សំរាយបញ្ជាក់

តាមវិសមភាព Holder ចំពោះចំនួនវិជ្ជមាន a, b, c គេបាន

\displaystyle \biggl(\frac{a^3}{b^3}+\frac{b^3}{c^3}+\frac{c^3}{a^3}\biggl)(\frac{a^3}{b^3}+\frac{b^3}{c^3}+\frac{c^3}{a^3}\biggl)\biggl(1^3+1^3+1^3\biggl)\geq \biggl(\frac{a}{b}.\frac{a}{b}.1+\frac{b}{c}.\frac{b}{c}.1+\frac{c}{a}.\frac{c}{a}.1\biggl)^3

\displaystyle 3\biggl(\frac{a^3}{b^3}+\frac{b^3}{c^3}+\frac{c^3}{a^3}\biggl)^2\geq \biggl(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\biggl)^3 ; (I)

ដោយ a\leq b\leq c\Longrightarrow a^2\leq b^2\leq c^2

តាង \displaystyle f(x)=\frac{1}{x} , x>0

\displaystyle f'(x)=-\frac{1}{x^2}

\displaystyle f''(x)=\frac{2}{x^3}>0 ; \forall{x>0}

តាមវិសមភាព V – K ចំពោះ a^2\leq b^2\leq c^2 និង f''(x)>0 គេបាន

(c^2-b^2)f(a^2)-(c^2-a^2)f(b^2)+(b^2-a^2)f(c^2)\geq 0

\displaystyle \frac{c^2-b^2}{a^2}-\frac{c^2-a^2}{b^2}+\frac{b^2-a^2}{c^2}\geq 0

\displaystyle \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\geq \frac{b^2}{a^2}+\frac{c^2}{b^2}+\frac{a^2}{c^2}

\displaystyle \biggl(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\biggl)^3\geq \biggl(\frac{b^2}{a^2}+\frac{c^2}{b^2}+\frac{a^2}{c^2}\biggl)^3  ; (II)

តាម (I) និង (II) គេបាន

\displaystyle 3\biggl(\frac{a^3}{b^3}+\frac{b^3}{c^3}+\frac{c^3}{a^3}\biggl)^2\geq \biggl(\frac{b^2}{a^2}+\frac{c^2}{b^2}+\frac{a^2}{c^2}\biggl)^3

លំហាត់ទើបច្នៃ

គេអោយបីចំនួនវិជ្ជមាន a, b, c ដែល a\leq b\leq c ។ ស្រាយបញ្ជាក់ថា

\displaystyle \frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\geq \frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}

សំរាយបញ្ជាក់

យើងមាន a\leq b\leq c\Longrightarrow a^3\leq b^3\leq c^3

តាង \displaystyle f(x)=\frac{1}{\sqrt[3]{x^2}} , x>0

\displaystyle f'(x)=-\frac{2}{3\sqrt[3]{x^5}}

\displaystyle f''(x)=\frac{10}{9\sqrt[3]{x^8}}>0 , \forall{x>0}

តាមវិសមភាព V – K ចំពោះ a^3\leq b^3\leq c^3 និង f''(x)>0 គេបាន

(c^3-b^3)f(a^3)-(c^3-a^3)f(b^3)+(b^3-a^3)f(c^3)\geq 0

\displaystyle \frac{c^3-b^3}{\sqrt[3]{a^6}}-\frac{c^3-a^3}{\sqrt[3]{b^6}}+\frac{b^3-a^3}{\sqrt[3]{c^6}}\geq 0

\displaystyle \frac{c^3-b^3}{a^2}-\frac{c^3-a^3}{b^2}+\frac{b^3-a^3}{c^2}\geq 0

\displaystyle \frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\geq \frac{b^3}{a^2}+\frac{c^3}{b^2}+\frac{a^3}{c^2} ; (\alpha )

យើងមាន

\displaystyle \biggl(\frac{b^2}{\sqrt{a^2}}+\frac{c^2}{\sqrt{b^2}}+\frac{a^2}{\sqrt{c^2}}\biggl)\biggl(\sqrt{a^2}+\sqrt{b^2}+\sqrt{c^2}\biggl)\geq (a+b+c)^2

\displaystyle \biggl(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\biggl)(a+b+c)\geq (a+b+c)^2 ; (I)

\displaystyle \biggl(\frac{b^4}{a^2\sqrt{b^2}}+\frac{c^4}{b^2\sqrt{c^2}}+\frac{a^4}{c^2\sqrt{a^2}}\biggl)\biggl(\sqrt{a^2}+\sqrt{b^2}+\sqrt{c^2}\biggl)\geq \biggl(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\biggl)^2

\displaystyle \Longrightarrow \biggl(\frac{b^3}{a^2}+\frac{c^3}{b^2}+\frac{a^3}{c^2}\biggl)(a+b+c)\geq \biggl(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\biggl)^2 ; (II)

គុណ (I) នឹង (II) គេបាន

\displaystyle \biggl(\frac{b^3}{a^2}+\frac{c^3}{b^2}+\frac{a^3}{c^2}\biggl)\biggl(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\biggl)\biggl(a+b+c)^2 \displaystyle \geq \biggl(a+b+c)^2\biggl(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\biggl)^2

\displaystyle \frac{b^3}{a^2}+\frac{c^3}{b^2}+\frac{a^3}{c^2}\geq \frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c} ; (\beta )

តាម (\alpha ) និង (\beta ) គេបាន

\displaystyle \frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\geq \frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}

ដូចនេះវិសមភាពត្រូវបានស្រាយបញ្ជាក់ 😀