Nice two degree inequality (1)--from amparvardi

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Nice two degree inequality (1)--from amparvardi

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#1 h Sep 16, 2010, 1:11 AM • 2 Y

Y by Adventure10, Mango247

see:http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=366988
Let $x_1,x_2,\cdots,x_n\ge0,n\ge2,$ prove that

$(x_1+x_2+\cdots+x_n)^2\ge{4(x_1x_2+x_2x_3+\cdots+x_{n-1}x_n)}$

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arqady

#2 h Sep 16, 2010, 3:45 AM • 2 Y

Y by Adventure10, Mango247

fjwxcsl wrote:

see:http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=366988
Let $x_1,x_2,\cdots,x_n\ge0,n\ge2,$ prove that

$(x_1+x_2+\cdots+x_n)^2\ge{4(x_1x_2+x_2x_3+\cdots+x_{n-1}x_n)}$

Because $(x_1+x_2+\cdots+x_n)^2\ge{4(x_1x_2+x_2x_3+\cdots+x_{n-1}x_n+x_nx_1)}$ .

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#3 h Sep 16, 2010, 4:15 AM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

fjwxcsl wrote:

see:http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=366988
Let $x_1,x_2,\cdots,x_n\ge0,n\ge2,$ prove that

$(x_1+x_2+\cdots+x_n)^2\ge{4(x_1x_2+x_2x_3+\cdots+x_{n-1}x_n)}$

Because $(x_1+x_2+\cdots+x_n)^2\ge{4(x_1x_2+x_2x_3+\cdots+x_{n-1}x_n+x_nx_1)}$ .

Try $n=2,3.$

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kuing

#4 h Sep 16, 2010, 5:24 AM • 2 Y

Y by Adventure10, Mango247

fjwxcsl wrote:

arqady wrote:

fjwxcsl wrote:

see:http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=366988
Let $x_1,x_2,\cdots,x_n\ge0,n\ge2,$ prove that

$(x_1+x_2+\cdots+x_n)^2\ge{4(x_1x_2+x_2x_3+\cdots+x_{n-1}x_n)}$

Because $(x_1+x_2+\cdots+x_n)^2\ge{4(x_1x_2+x_2x_3+\cdots+x_{n-1}x_n+x_nx_1)}$ .

Try $n=2,3.$

that inequality holds for $n\ge4$

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kuing

#5 h Sep 16, 2010, 7:56 AM • 1 Y

Y by Adventure10

$n \ge 4,x_1 ,x_2 ,...,x_n \ge 0$ , then $(x_1 + x_2 + \cdots + x_n )^2 \ge 4(x_1 x_2 + x_2 x_3 + \cdots + x_{n - 1} x_n + x_n x_1 ).$
use induction to prove.

when $n=4$ , equivalent to

$(x_1 + x_2 + x_3 + x_4 )^2 \ge 4(x_1 x_2 + x_2 x_3 + x_3 x_4 + x_4 x_1 )$

$\Leftrightarrow (x_1 + x_2 + x_3 + x_4 )^2 \ge 4\left( {x_1 + x_3 } \right)\left( {x_2 + x_4 } \right),$

obvious true by Am-Gm.

assume when $n=k$ inequality holds, i.e.

$(x_1 + x_2 + \cdots + x_k )^2 \ge 4(x_1 x_2 + x_2 x_3 + \cdots + x_{k - 1} x_k + x_k x_1 ),$

then when $n=k+1$ , WOLG, assume $x_1 = \max \left\{ {x_1 ,x_2 ,...,x_n } \right\}$ , we have

$(x_1 + x_2 + \cdots + x_k + x_{k + 1} )^2$

$\ge 4(x_1 x_2 + x_2 x_3 + \cdots + x_{k - 1} ( {x_k + x_{k + 1} } ) + ( {x_k + x_{k + 1} } )x_1 )$

$= 4(x_1 x_2 + x_2 x_3 + \cdots + x_{k - 1} x_k + x_k x_{k + 1} + x_{k + 1} x_1 + x_{k - 1} x_{k + 1} + x_1 x_k - x_k x_{k + 1} )$

$\ge 4(x_1 x_2 + x_2 x_3 + \cdots + x_{k - 1} x_k + x_k x_{k + 1} + x_{k + 1} x_1 ).$

so $n=k+1$ is also true, the prove is completed.

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