Algebra: Inequalities
Let a, b, c be real numbers. Prove the inequalities:
a) $a^2 + b^2 + c^2 \geq ab + bc + ca$;
b) $a^4 + b^4 + c^4 \geq abc(a + b + c)$.
a) $a^2 + b^2 + c^2 \geq ab + bc + ca$;
b) $a^4 + b^4 + c^4 \geq abc(a + b + c)$.
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- Nadim Ul Abrar
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Re: Algebra: Inequalities
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
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
$\frac{1}{0}$
- nafistiham
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Re: Algebra: Inequalities
here,
a)
a)
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: Algebra: Inequalities
again,
b)
b)
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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- Nadim Ul Abrar
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Re: Algebra: Inequalities
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
$\frac{x^{2}+y^{2}}{2} \geq xy$*
$\frac{y^{2}+z^{2}}{2} \geq yz$*
$\frac{z^{2}+x^{2}}{2} \geq zx$*
added three stars
$\frac{1}{0}$
Re: Algebra: Inequalities
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
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Re: Algebra: Inequalities
@ willpower : It's because of re-arrangement inequality.
@nafistiham
@nafistiham
How?nafistiham wrote:again,
\[\Rightarrow a^{4}+b^{4}+c^{4} \geq a^{2}bc+b^{2}ca+c^{2}ab\]
You spin my head right round right round,
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
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Re: Algebra: Inequalities
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
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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Re: Algebra: Inequalities
আমি এটাকে rearrangement inequality ভেবেছিলাম। ভুলটা বুঝতে পেরেছি । ধন্যবাদ। ঠিকমত সমাধান করার পর এডিট করে দেব।sourav das wrote:@ willpower : It's because of re-arrangement inequality.
@nafistihamHow?nafistiham wrote:again,
\[\Rightarrow a^{4}+b^{4}+c^{4} \geq a^{2}bc+b^{2}ca+c^{2}ab\]
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
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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Re: Algebra: Inequalities
Reply please. (Because according to re-arrangement inequality you can't explain it. So...)sourav das wrote:@ willpower : It's because of re-arrangement inequality.
@nafistihamHow?nafistiham wrote:again,
\[\Rightarrow a^{4}+b^{4}+c^{4} \geq a^{2}bc+b^{2}ca+c^{2}ab\]
You spin my head right round right round,
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )