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Ari004

#1 h Sep 15, 2018, 12:04 PM • 4 Y

Y by Pointcircle, BarisKoyuncu, Adventure10, Mango247

Let $a,b,c$ be positive real numbers, such that: $a+b+c \ge \frac {1}{a}+\frac {1}{b}+\frac{1}{c}$ .

Prove that:
$a+b+c \ge \frac{3}{a+b+c}+\frac {2}{abc}$ .

This post has been edited 1 time. Last edited by Ari004, Sep 15, 2018, 12:05 PM

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#2 h Sep 15, 2018, 12:22 PM • 3 Y

Y by Pointcircle, Adventure10, Mango247

C-S helps here
We have $abc(a+b+c)\geq ab+bc+ca.$
. We must prove that: $abc(a+b+c)^{2}\geq3abc+2(a+b+c).$
We have that: $abc(a+b+c)^{2}\geq(a+b+c)(ab+bc+ca)=3abc+ab(a+b)+bc(b+c)+ca(c+a).$
It is enought to prove that $ab(a+b)+bc(b+c)+ca(c+a)\geq2(a+b+c)$
The last easy by C-S

This post has been edited 1 time. Last edited by unrulykid3, Sep 15, 2018, 12:24 PM
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#3 h Nov 27, 2020, 6:08 AM • 1 Y

Y by Mango247

unrulykid3 wrote:

C-S helps here
We have $abc(a+b+c)\geq ab+bc+ca.$
. We must prove that: $abc(a+b+c)^{2}\geq3abc+2(a+b+c).$
We have that: $abc(a+b+c)^{2}\geq(a+b+c)(ab+bc+ca)=3abc+ab(a+b)+bc(b+c)+ca(c+a).$
It is enought to prove that $ab(a+b)+bc(b+c)+ca(c+a)\geq2(a+b+c)$
The last easy by C-S

How can it be deduced by Cauchy-Schwarz?

This post has been edited 1 time. Last edited by Universes, Nov 27, 2020, 6:09 AM

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hyay

#4 h Nov 27, 2020, 7:36 AM

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$2(ab(a+b) + bc(b+c) + ca(c+a))(a + b + c) \geq (ab(a+b) + bc(b+c) + ca(c+a))(\frac{2}{a} + \frac{2}{b} + \frac{2}{c})$
$= (ab(a+b) + bc(b+c) + ca(c+a))(\frac{a+b}{ab} + \frac{b+c}{bc} + \frac{ca}{c+a}) \geq 4(a + b + c)^2$

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