Let a,b,c are positive real numbers$(a+b+c)(\dfrac{1}{a})=10

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Let a,b,c are positive real numbers$(a+b+c)(\dfrac{1}{a})=10

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duykhuongvietnam

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duykhuongvietnam

#1 h Aug 24, 2015, 7:55 AM • 2 Y

Y by Adventure10, Mango247

Let a,b,c are positive real numbers such that: $(a+b+c)(\sum\dfrac{1}{a})=10$
Find the Min and Max of P= $(\sum a^2)(\sum \dfrac{1}{a^2})$

This post has been edited 2 times. Last edited by duykhuongvietnam, Aug 24, 2015, 7:56 AM

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41534 posts

sqing

#2 h Aug 24, 2015, 9:07 AM • 2 Y

Y by Adventure10, Mango247

Vasc:
If $a,b,c$ are positive real numbers such that $(a+b+c)\left(\frac 1{a}+\frac 1{b}+\frac 1{c}\right)=10,$
then $\frac {19}{12}\le \frac a{b+c}+\frac b{c+a}+\frac c{a+b}\le \frac 5{3};$ $\frac {69}{40}\le \frac {a^2}{b^2+c^2}+\frac {b^2}{c^2+a^2}+\frac {c^2}{a^2+b^2}\le \frac {12}{5}.$

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math90

#3 h Aug 24, 2015, 9:22 AM • 2 Y

Y by Adventure10, Mango247

For Min:
Set $x=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ , $y=\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$
The condition becomes $x+y=7$ .
We also have
$P=x^2+y^2-2x-2y+3=1+\frac{1}{2}[(x+y-2)^2+(x-y)^2]\geq1+\frac{1}{2}(7-2)^2=13.5$
Thus, the minimum is $13.5$ .
Equality holds when $a=b=\frac{1}{2}c$ , or any cyclic permutation.

For Max:
Using the known inequality
$\sqrt{(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}\geq 1+\sqrt{1+\sqrt{(a^2+b^2+c^2)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)}}$

We get that
$P\leq 140-40\sqrt{10}$
Equality holds when $a^2=bc$ and $\frac{b}{c}=\frac{(3+\sqrt{5})(3-2\sqrt{2})}{2}$ , or any symmetric permutation.

This post has been edited 1 time. Last edited by math90, Aug 24, 2015, 9:25 AM

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sqing

#7 h Oct 16, 2017, 1:52 AM • 1 Y

Y by Adventure10

duykhuongvietnam wrote:

Let a,b,c are positive real numbers such that: $(a+b+c)(\sum\dfrac{1}{a})=10$
Find the Min and Max of P= $(\sum a^2)(\sum \dfrac{1}{a^2})$

Saudi Arabia Mathematical Competitions 2015
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