With $a,b,c\ge 0$

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#1 h Sep 24, 2017, 12:52 PM • 2 Y

Y by Adventure10, Mango247

If $a,b,c$ are nonnegative real numbers, then
$\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{4ab}{(a+b)^2}\ge 2$

This post has been edited 1 time. Last edited by khanhsy, Sep 24, 2017, 2:02 PM
Reason: edit

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dipdas

#2 h Sep 24, 2017, 1:03 PM • 2 Y

Y by Adventure10, Mango247

$(a^2+b^2+c^2)(b^2+c^2+a^2){\geq}(ab+bc+ca)^2$ by CS inequality
$=>\frac{a^2+b^2+c^2}{ab+bc+ca}{\geq}1$ .......(i)

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dipdas

#3 h Sep 24, 2017, 1:15 PM • 2 Y

Y by Adventure10, Mango247

khanhsy wrote:

If $a,b,c$ are nonnegative real numbers, then
$\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{27abc}{(a+b+c)^3}\ge 2$

is the inequality correct?
take a=1,b=c=0

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pym123

#4 h Sep 24, 2017, 1:17 PM • 2 Y

Y by Adventure10, Mango247

How could you divide a number with 0?

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dipdas

#5 h Sep 24, 2017, 1:18 PM • 2 Y

Y by Adventure10, Mango247

pym123 wrote:

How could you divide a number with 0?

so your inequality is not correct since the numbers are non negative

This post has been edited 1 time. Last edited by dipdas, Sep 24, 2017, 1:19 PM

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#6 h Sep 24, 2017, 1:21 PM • 1 Y

Y by Adventure10

I guess the OP means that no two of $a,b,c$ are zero.

By the way

dipdas wrote:

$(a^2+b^2+c^2)(b^2+c^2+a^2){\geq}(ab+bc+ca)^2$ by CS inequality
$=>\frac{a^2+b^2+c^2}{ab+bc+ca}{\geq}1$ .......(i)

this approach doesn't help.

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dipdas

#7 h Sep 24, 2017, 1:22 PM • 1 Y

Y by Adventure10

knm2608 wrote:

By the way

dipdas wrote:

$(a^2+b^2+c^2)(b^2+c^2+a^2){\geq}(ab+bc+ca)^2$ by CS inequality
$=>\frac{a^2+b^2+c^2}{ab+bc+ca}{\geq}1$ .......(i)

this approach doesn't help.

i know

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pym123

#9 h Sep 24, 2017, 1:26 PM • 2 Y

Y by Adventure10, Mango247

The inequality is equivalent to
$\sum a^5 + 3\sum a^4b +3\sum ab^4 + 9\sum a^2b^2c \ge 4\sum a^3b^2 + 4\sum a^2b^3 + 12\sum a^3bc$

This post has been edited 1 time. Last edited by pym123, Sep 24, 2017, 1:29 PM
Reason: typo

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#10 h Sep 24, 2017, 1:30 PM • 2 Y

Y by Adventure10, Mango247

By $uvw$ we only have to check two cases.

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#11 h Sep 24, 2017, 1:50 PM • 2 Y

Y by Adventure10, Mango247

Sorry. That is no correct?

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dipdas

#12 h Sep 24, 2017, 2:06 PM • 3 Y

Y by khanhsy, Adventure10, Mango247

khanhsy wrote:

Sorry. That is no correct?

i already told you

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#13 h Sep 24, 2017, 2:44 PM • 2 Y

Y by Adventure10, Mango247

dipdas wrote:

khanhsy wrote:

Sorry. That is no correct?

i already told you

Thank you, Dipdas! I have fixed my post.

This post has been edited 1 time. Last edited by khanhsy, Sep 24, 2017, 2:45 PM
Reason: .

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pym123

#14 h Sep 24, 2017, 3:35 PM • 2 Y

Y by khanhsy, Adventure10

This inequality is equivalent to
$a^4 + 2a^2b^2 + b^4 + a^2c^2 + 2abc^2 +b^2c^2 \ge 2a^3c + 2abc^2 + 2b^3c + 2ab^2c$
And it is equivalent to $(a^2 + b^2)^2 + c^2(a + b)^2 \ge 2c(a^2 + b^2)(a + b)$
Which is correct by AM - GM.

This post has been edited 1 time. Last edited by pym123, Sep 24, 2017, 3:35 PM

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