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inequality

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Nguyễn Huy Thắng
09/03/2017 at 20:58
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Let \(a,b,c\) are positive real number satisfy \(a+b+c=1\). Prove that \(\dfrac{1}{\sqrt{\left (a^2+ab+b^2\right )\left (b^2+bc+c^2\right )}}+\dfrac{1}{\sqrt{\left (b^2+bc+c^2\right )\left (c^2+ca+a^2\right )}}+\dfrac{1}{\sqrt{\left (c^2+ca+a^2\right )\left (a^2+ab+b^2\right )}}\ge 4+\dfrac{8}{\sqrt{3}}\)

inequality

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    Nguyen Huu Ai Linh 10/12/2017 at 07:08

    That too long!


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Nguyễn Nhật Minh
04/04/2017 at 18:29
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Prove that:

B = \(\dfrac{1}{2!}+\dfrac{5}{3!}+\dfrac{11}{4!}+...+\dfrac{n^2+n-1}{\left(n+1\right)!}< 2\).

Help me please! This is my homework and I must give it to my teacher tomorrow!

inequalityFraction


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Nguyễn Nhật Minh
02/04/2017 at 11:25
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Prove that E = \(\dfrac{1}{3^3}+\dfrac{1}{4^3}+\dfrac{1}{5^3}+...+\dfrac{1}{n^3}< \dfrac{1}{12}\).

Fractioninequality

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    FA KAKALOTS 28/01/2018 at 22:08

    For any natural number n > 1,we have :

    (n - 1)n(n + 1) = n(n2 - 1) = n3 - n < n3

    ⇒1n3<1(n−1)n(n+1)

    1(n−1)n(n+1)=1n.1(n−1)(n+1)

    =1n.(n+1)−(n−1)(n−1)(n+1).12=12.1n.(1n−1−1n+1)

    =12.(1(n−1)n−1n(n+1))

    Now we have :

    E < 12.3.4+13.4.5+14.5.6+...+1(n−1)n(n+1)

    =12(12.3−13.4)+12(13.4−14.5)+12(14.5−15.6)+...+12(1(n−1)n−1n(n+1))

    =12(12.3−1n(n+1))=112−12n(n+1)<112

    Hence,E<112

  • ...
    Phan Thanh Tinh Coordinator 24/04/2017 at 13:50

    For any natural number n > 1,we have :

    (n - 1)n(n + 1) = n(n2 - 1) = n3 - n < n3

    \(\Rightarrow\dfrac{1}{n^3}< \dfrac{1}{\left(n-1\right)n\left(n+1\right)}\)

    \(\dfrac{1}{\left(n-1\right)n\left(n+1\right)}=\dfrac{1}{n}.\dfrac{1}{\left(n-1\right)\left(n+1\right)}\)

    \(=\dfrac{1}{n}.\dfrac{\left(n+1\right)-\left(n-1\right)}{\left(n-1\right)\left(n+1\right)}.\dfrac{1}{2}=\dfrac{1}{2}.\dfrac{1}{n}.\left(\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)\)

    \(=\dfrac{1}{2}.\left(\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\right)\)

    Now we have :

    E < \(\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+\dfrac{1}{4.5.6}+...+\dfrac{1}{\left(n-1\right)n\left(n+1\right)}\)

    \(=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)+\dfrac{1}{2}\left(\dfrac{1}{3.4}-\dfrac{1}{4.5}\right)+\dfrac{1}{2}\left(\dfrac{1}{4.5}-\dfrac{1}{5.6}\right)+...+\dfrac{1}{2}\left(\dfrac{1}{\left(n-1\right)n}-\dfrac{1}{n\left(n+1\right)}\right)\)

    \(=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{n\left(n+1\right)}\right)=\dfrac{1}{12}-\dfrac{1}{2n\left(n+1\right)}< \dfrac{1}{12}\)

    Hence,\(E< \dfrac{1}{12}\)


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