Solve this PLEASE!

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Prosenjit Basak
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Solve this PLEASE!

Unread post by Prosenjit Basak » Fri Mar 22, 2013 10:37 pm

If $a , b , c > 0$, prove that
$\displaystyle \frac{9}{a + b + c} \leq 2(\frac{1}{a + b} + \frac{1}{b + c} + \frac{1}{c + a}) \leq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$
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Phlembac Adib Hasan
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Re: Solve this PLEASE!

Unread post by Phlembac Adib Hasan » Sat Mar 23, 2013 10:02 am

Prosenjit Basak wrote:If $a , b , c > 0$, prove that
$\displaystyle \frac{9}{a + b + c} \leq 2(\frac{1}{a + b} + \frac{1}{b + c} + \frac{1}{c + a}) \leq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$
From Cauchy-Schwarz's inequality, we know that \[ \left (\frac {a_1^2}{b_1}+\frac {a_2^2}{b_2}+...+\frac {a_n^2}{b_n}\right )(b_1+b_2+...+b_n)\ge (a_1+a_2+...+a_n)^2\]
Using this we derive both of the inequalities: \[\frac{1^2}{a + b} + \frac{1^2}{b + c} + \frac{1^2}{c + a}\ge \frac {(1+1+1)^2}{(a+b)+(b+c)+(c+a)}=\frac {9}{2(a+b+c)}\]
\[\left.\begin {array}{lr}\displaystyle \frac 1 a+\frac 1 b\ge \frac 4 {a+b}\\
\frac 1 b+\frac 1 c\ge \frac 4 {b+c}\\
\frac 1 c+\frac 1 a\ge \frac 4 {c+a}\end{array}
\right \} 2\left (\frac 1 a+\frac 1 b+\frac 1 c\right )\ge 4\left (\frac 1 {a+b}+\frac 1 {b+c}+\frac 1 {c+a}\right )\]
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SANZEED
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Re: Solve this PLEASE!

Unread post by SANZEED » Sat Mar 23, 2013 11:09 pm

Prosenjit Basak wrote:If $a , b , c > 0$, prove that
$\displaystyle \frac{9}{a + b + c} \leq 2(\frac{1}{a + b} + \frac{1}{b + c} + \frac{1}{c + a}) \leq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$
According to the AM-HM inequality,we know that,for positive real numbers $x_1,x_2,...,x_n$,
$\displaystyle\frac{n}{\frac{1}{x_1}+\frac{1}{x_2}+...+\frac{1}{x_n}}\leq \frac{x_1+x_2+...+x_n}{n}$.
Now apply this inequality for $x_1=\frac{2}{a+b},x_2=\frac{2}{b+c},x_3=\frac{2}{c+a}$ to get that
$\displaystyle\frac{1}{3}\cdot (\displaystyle\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}$
$\displaystyle\geq \frac{3}{\displaystyle\frac{1}{\frac{2}{a+b}}+\frac{1}{\frac{2}{b+c}}+\frac{1}{\frac{2}{c+a}}}$
$\displaystyle=\frac{3}{\displaystyle\frac{a+b}{2}+\frac{b+c}{2}+\frac{c+a}{2}}$
$\displaystyle=\frac{3}{a+b+c}$
For the remaining inequality use AM-HM inequality for $(\frac{1}{a},\frac{1}{b}),(\frac{1}{b},\frac{1}{c}),(\frac{1}{c},\frac{1}{a})$ and add the three inequalities together.
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Prosenjit Basak
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Joined:Wed Nov 28, 2012 12:48 pm

Re: Solve this PLEASE!

Unread post by Prosenjit Basak » Mon Mar 25, 2013 10:26 pm

HM মানে হারমনিক মিন না? তাইলে ঠিক আছে । আর সানজিদ ভাই এর সলিউশনটা সহজ ছিল। আসলে আমি Cauchy-Schwarz's inequality-টা কি তা জানি না তাই প্রবলেম হইছে।
Last edited by Phlembac Adib Hasan on Tue Mar 26, 2013 12:22 pm, edited 1 time in total.
Reason: Fixing typo
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Phlembac Adib Hasan
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Re: Solve this PLEASE!

Unread post by Phlembac Adib Hasan » Tue Mar 26, 2013 12:21 pm

Prosenjit Basak wrote:HM মানে হারমনিক মিন না? তাইলে ঠিক আছে । আর সানজিদ ভাই এর সলিউশনটা সহজ ছিল। আসলে আমি Cauchy-Schwarz's inequality-টা কি তা জানি না তাই প্রবলেম হইছে।
সত্যিকারের কশি বলতে গেলে আমাদের কেউই জানে না। :) $\;$ আসল কশি ভেক্টর স্পেসের উপর ডিফাইন করা হয়েছে। আমি বা আমরা যেটা ব্যবহার করি সেটা মূল কশির একটা স্পেশাল কেস মাত্র।
(আমাদের যেটা লাগে) কশির inequality বা Cauchy-Schwarz's Inequality বা Cauchy-Schwarz-Bunyakovsky's Inequality হচ্ছে,
If $a_1, a_2,..., a_n$ are some non-zero real numbers and if $b_1, b_2,..., b_n$ are some more non-zero real numbers, then this inequality holds:
\[(a_1^2+a_2^2+...+a_n^2)(b_1^2+b_2^2+...+b_n^2)\geq (a_1b_1+a_2b_2+...+a_nb_n)^2\]
with equality iff $\displaystyle \frac {a_1}{b_1}=\frac {a_2}{b_2}=...=\frac {a_n}{b_n}=$ some constant.

Prosenjit Basak
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Re: Solve this PLEASE!

Unread post by Prosenjit Basak » Wed Mar 27, 2013 9:02 pm

যেইদিন পোষ্ট করলাম সেইদিনই রাতে একখান বইয়ে দেখি আরে এইতো Cauchy - schwarz inequality. যদিও ঐখানে এত সহজ কইরা দেয় নাই
একটু কটমটে ইকুয়েশন দিছিল। যাই হোক পরে বুঝছি।
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