Algebra: Inequalities

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willpower
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Algebra: Inequalities

Unread post by willpower » Tue Nov 01, 2011 9:16 pm

Let a, b, c be real numbers. Prove the inequalities:
a) $a^2 + b^2 + c^2 \geq ab + bc + ca$;
b) $a^4 + b^4 + c^4 \geq abc(a + b + c)$.
Everybody is a genius; but if you judge a fish on its ability to climb a tree, it will live its entire life believing that it is stupid. - Albert Einstein

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Nadim Ul Abrar
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Re: Algebra: Inequalities

Unread post by Nadim Ul Abrar » Tue Nov 01, 2011 9:29 pm

a) rearrangement .
b)

$a^{4}+b^{4}+c^{4}=3[4,0,0]$

$abc(a+b+c)=3[2,1,1]$

$[4,0,0] \geq 3[2,1,1]$

so proved :D
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nafistiham
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Re: Algebra: Inequalities

Unread post by nafistiham » Tue Nov 01, 2011 9:30 pm

here,
a)
\[a^{2}+b^{2}+c^{2}\geq ab+bc+ca\]
\[2a^{2}+2b^{2}+2c^{2}\geq2ab+2bc+2ca\]
\[a^{2}-2ab+b^{2}+b^{2}-2bc+c^{2}+c^{2}-2ca+a^{2}\]
\[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\geq0\]
which is true
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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nafistiham
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Re: Algebra: Inequalities

Unread post by nafistiham » Tue Nov 01, 2011 9:35 pm

again,
b)
suppose,
\[a \geq b \geq c\]
by rearrangements,
\[a+b+c=a+b+c\]
\[\Rightarrow a^{2}+b^{2}+c^{2}=a^{2}+b^{2}+c^{2}\]
\[\Rightarrow a^{3}+b^{3}+c^{3} \geq a^{2}b+b^{2}c+c^{2}a\]
\[\Rightarrow a^{4}+b^{4}+c^{4} \geq a^{2}bc+b^{2}ca+c^{2}ab\]
\[\Rightarrow a^{4}+b^{4}+c^{4} \geq abc(a+b+c)\]
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Introduction:
Nafis Tiham
CSE Dept. SUST -HSC 14'
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Nadim Ul Abrar
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Re: Algebra: Inequalities

Unread post by Nadim Ul Abrar » Tue Nov 01, 2011 9:40 pm

here is a)
$\frac{x^{2}+y^{2}}{2} \geq xy$*

$\frac{y^{2}+z^{2}}{2} \geq yz$*

$\frac{z^{2}+x^{2}}{2} \geq zx$*

added three stars
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willpower
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Re: Algebra: Inequalities

Unread post by willpower » Tue Nov 01, 2011 10:07 pm

What is the reason behind this step? \[a^3 + b^3 + c^3 \geq a^2b + b^2c + c^2a\]
Everybody is a genius; but if you judge a fish on its ability to climb a tree, it will live its entire life believing that it is stupid. - Albert Einstein

sourav das
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Re: Algebra: Inequalities

Unread post by sourav das » Tue Nov 01, 2011 10:15 pm

@ willpower : It's because of re-arrangement inequality.
@nafistiham
nafistiham wrote:again,
\[\Rightarrow a^{4}+b^{4}+c^{4} \geq a^{2}bc+b^{2}ca+c^{2}ab\]
How?
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

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nafistiham
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Re: Algebra: Inequalities

Unread post by nafistiham » Tue Nov 01, 2011 10:24 pm

see this

http://www.matholympiad.org.bd/forum/vi ... =14&t=1306

or, for more information see this book

ifile.it/zlct48/ebooksclub.org__Inequalities__A_Mathematical_Olympiad_Approach.pdf
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
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nafistiham
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Re: Algebra: Inequalities

Unread post by nafistiham » Tue Nov 01, 2011 10:27 pm

sourav das wrote:@ willpower : It's because of re-arrangement inequality.
@nafistiham
nafistiham wrote:again,
\[\Rightarrow a^{4}+b^{4}+c^{4} \geq a^{2}bc+b^{2}ca+c^{2}ab\]
How?
আমি এটাকে rearrangement inequality ভেবেছিলাম। ভুলটা বুঝতে পেরেছি । ধন্যবাদ। ঠিকমত সমাধান করার পর এডিট করে দেব।
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Introduction:
Nafis Tiham
CSE Dept. SUST -HSC 14'
http://www.facebook.com/nafistiham
nafistiham@gmail

sourav das
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Re: Algebra: Inequalities

Unread post by sourav das » Tue Nov 01, 2011 10:32 pm

sourav das wrote:@ willpower : It's because of re-arrangement inequality.
@nafistiham
nafistiham wrote:again,
\[\Rightarrow a^{4}+b^{4}+c^{4} \geq a^{2}bc+b^{2}ca+c^{2}ab\]
How?
Reply please. (Because according to re-arrangement inequality you can't explain it. So...)
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

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