3 variable with fraction inequality

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3 variable with fraction inequality

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Jonru

#1 h Jun 26, 2016, 11:42 AM • 1 Y

Y by Adventure10

For integers $a,b,c>0$ prove that:
$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge \frac{3}{2} + \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$

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Reynan

#2 h Jun 26, 2016, 12:12 PM • 4 Y

Y by Jonru, CaptainCuong, Adventure10, Mango247

$\sum \frac{a}{b}-\frac{a}{b+c}=\sum \frac{ac}{b(b+c)}$
let $a=\frac{1}{x},b=\frac{1}{y},c=\frac{1}{z}$
$\sum \frac{ac}{b(b+c)} = \sum \frac{y^2}{xz+xy}\ge \frac{(x+y+z)^2}{2(xy+yz+zx)}\ge \frac{3}{2}$

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#3 h Jun 27, 2016, 1:41 AM • 1 Y

Y by Adventure10

Jonru wrote:

For integers $a,b,c>0$ prove that:
$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge \frac{3}{2} + \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$

Remark:
Use that problem, we have if a, b, c>0 such that: $ab + bc + ca + 2abc = 1$ . Prove that: $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge 3 + 2\left( {a + b + c} \right)$ .

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Reynan

#4 h Jun 27, 2016, 10:01 AM • 3 Y

Y by CaptainCuong, Adventure10, Mango247

$ab + bc + ca + 2abc = 1$ let $a=\frac{x}{y+z},b=\frac{y}{z+x},c=\frac{z}{x+y}$
$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge 3 + 2\left( {a + b + c} \right)\Leftrightarrow \frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\ge 3 +\frac{2x}{y+z}+\frac{2y}{z+x}+\frac{2z}{x+y}$
$\sum \frac{y}{x}+\frac{y}{z}-\frac{2y}{x+z}\ge 3$
$\sum \frac{y}{x}+\frac{y}{z}-\frac{2y}{x+z}=\sum \frac{y(x^2+z^2)}{xz(x+y)}$ let $x=\frac{1}{p},y=\frac{1}{q},z=\frac{1}{r}$
$\sum \frac{y(x^2+z^2)}{xz(x+y)}=\sum \frac{p^2+r^2}{pq+qr}\ge \sum \frac{2pr}{pq+pr}\ge 3$ this is nesbit

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arqady

#6 h Aug 21, 2016, 6:28 AM • 2 Y

Y by Adventure10, Mango247

Jonru wrote:

For integers $a,b,c>0$ prove that:
$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge \frac{3}{2} + \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$

It was here:
http://www.artofproblemsolving.com/community/c6h98664

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sqing

#7 h Sep 21, 2024, 12:01 PM

Y by

Jonru wrote:

For integers $a,b,c>0$ prove that:
$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \ge \frac{3}{2} + \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b}$

Let $a,b,c>0.$ Prove that $\frac ab+\frac bc+\frac ca\geq \frac a{b+c}+\frac b{c+a}+\frac c{a+b}+\frac 32$ $\iff \frac{a}{b} + \frac{b}{c} + \frac{c}{a}+\frac{3}{2} \ge(a+b+c)( \frac{1}{a+b}+\frac{1}{b+c} + \frac{1}{c+a} )$ $\implies( \frac{a}{b} + \frac{b}{c} + \frac{c}{a})^2\ge 2(a+b+c)( \frac{1}{a+b}+\frac{1}{b+c} + \frac{1}{c+a} )$ Crux , (7) 2024,Q4970

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