Inequality

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  • #72385
    Akash Arjun
    Participant

     

    Please help me solve this problem

    #72538
    Saumik Karfa
    Moderator

    Using A.M. $\geq$ H.M.

    $\Rightarrow \frac{a+b+c}{3} \geq \frac{3}{\frac 1a +\frac 1b+\frac 1c}$

    $\Rightarrow (a+b+c)(\frac 1a +\frac 1b+\frac 1c) \geq 9$

    $\Rightarrow (a+b+c)^2 \geq 9$

    Let $A=a+b+c$

    then $A^2 \geq 9 \Rightarrow \frac{1}{A^2} \leq 9$

    let, $f(x)=\frac{1}{(A+x)^2}$

    $\lambda_i = \frac 13$

    $x_1=a,x_2=b,x_3=c$

    Function is concave.

    using JENSEN’s inequality :

    $\displaystyle\sum_{i=1}^n \lambda_i f(x_i) \leq f\bigg(\lambda_i x_1\bigg)$

    $\Rightarrow   \frac{\text{Required Expression}}{3}  \leq f\bigg(\frac A3\bigg)$

    $\Rightarrow   \frac{\text{Required Expression}}{3}  \leq \frac{1}{(A+\frac A3)^2}$

    $\Rightarrow  \frac{\text{Required Expression}}{3}  \leq \frac{9}{16A^2} \leq \frac {1}{16} $

    $\Rightarrow \text{ Required Expression } \leq \frac {3}{16} =\frac mn$

    Hence, $m+n=19$

     

     

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