help me at this pls i'm so confused

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High School Math Grades 9-12, Ages 13-18

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High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

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help me at this pls i'm so confused

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#1 h Jun 12, 2020, 2:36 PM

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if a, b, c are real numbers and a, b, c >0, prove that
pls help meee

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#2 h Jun 12, 2020, 2:49 PM • 1 Y

Y by agentP

is this from asmp??

This post has been edited 1 time. Last edited by srisainandan6, Jun 12, 2020, 2:57 PM

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#3 h Jun 12, 2020, 2:56 PM

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What is amsp?

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#4 h Jun 12, 2020, 2:57 PM • 1 Y

Y by agentP

awesome math summer program

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#5 h Jun 12, 2020, 2:58 PM • 1 Y

Y by Mango247

well i don't know what is amsp so i can't answer you this

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#6 h Jun 12, 2020, 3:00 PM • 1 Y

Y by agentP

ok when dealing with inequalities, try to use these basic ones first:

AM GM
Cauchy (Titu's lemma in specific cases)
trivial inequality

which one do you think can be used

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#7 h Jun 12, 2020, 3:11 PM

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Just trivial inequality here.

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#8 h Jun 12, 2020, 3:14 PM • 2 Y

Y by agentP, Mango247

yes, and from here you just want to algebraically manipulate this to get something in this form:
$(...)^2 + (...)^2 + .... + (....)^ 2$ $\ge$ $0$

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#9 h Jun 12, 2020, 3:27 PM

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This post has been edited 2 times. Last edited by stacy2020, Jun 12, 2020, 7:57 PM
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#10 h Jun 12, 2020, 3:40 PM

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Subtracting doesn't work like that for inequalities.
For example:
$5\geq4$
$3\geq -50$
Subtract to get $2\geq54$

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#12 h Jun 26, 2020, 6:06 AM • 1 Y

Y by Mango247

myname17042005 wrote:

if a, b, c are real numbers and a, b, c >0, prove that
pls help meee

my solution
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#13 h Jun 26, 2020, 6:53 AM • 1 Y

Y by Imayormaynotknowcalculus

myname17042005 wrote:

if a, b, c are real numbers and a, b, c >0, prove that
pls help meee
$\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\geq 2\left(\dfrac{a}{c}+\dfrac{b}{a}+\dfrac{c}{b}\right)-3$

It's equivalent to prove that: $x^2+y^2+z^2+3\geq 2(xy+yz+zx)$ , where $x,y,z>0$ such that $xyz=1$ .
Since $3=\dfrac{9xyz}{3\sqrt[3]{xyz}}\geq \dfrac{9xyz}{x+y+z}$ , it remains to show that:
$x^2+y^2+z^2+\dfrac{9xyz}{x+y+z}\geq 2(xy+yz+zx)$ which is Schur's inequality.

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sqing

#14 h Yesterday at 3:07 AM

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DoThinh2001 wrote:

myname17042005 wrote:

if a, b, c are real numbers and a, b, c >0, prove that
pls help meee
$\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\geq 2\left(\dfrac{a}{c}+\dfrac{b}{a}+\dfrac{c}{b}\right)-3$

It's equivalent to prove that: $x^2+y^2+z^2+3\geq 2(xy+yz+zx)$ , where $x,y,z>0$ such that $xyz=1$ .
Since $3=\dfrac{9xyz}{3\sqrt[3]{xyz}}\geq \dfrac{9xyz}{x+y+z}$ , it remains to show that:
$x^2+y^2+z^2+\dfrac{9xyz}{x+y+z}\geq 2(xy+yz+zx)$ which is Schur's inequality.

(MR J693)Let $a,b,c>0$ and $a+b+c=3abc .$ Prove that
$a^2+b^2+c^2+3\geq 2(ab+bc+ca)$ $\iff a^2+b^2+c^2+\dfrac{9abc}{a+b+c}\geq 2(ab+bc+ca)$ https://artofproblemsolving.com/community/c6h483075p9470528
https://artofproblemsolving.com/community/c6h1562485p9562400
https://artofproblemsolving.com/community/c6h37173p565225
https://artofproblemsolving.com/community/c6h76676p17106912
Let $a$ , $b$ and $c$ be positive numbers such that $a^2+b^2+c^2+1=4a^2b^2c^2$ . Prove that
$a^2+b^2+c^2+3\geq2(ab+ac+bc)$

This post has been edited 2 times. Last edited by sqing, Yesterday at 3:13 AM

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