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#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

prove :$\dfrac {ab+c}{c+1}+\dfrac {bc+a}{a+1}+\dfrac {ca+b}{b+1}\geq\dfrac {18}

{a+b+c+3}$

- Thread starter Albert
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- Thread starter
- #1

- Jan 25, 2013

- 1,225

prove :$\dfrac {ab+c}{c+1}+\dfrac {bc+a}{a+1}+\dfrac {ca+b}{b+1}\geq\dfrac {18}

{a+b+c+3}$

- Nov 26, 2013

- 719

which is obviously true for $a,b,c\geq1$

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- #3

- Jan 25, 2013

- 1,225

prove :$\dfrac {ab+c}{c+1}+\dfrac {bc+a}{a+1}+\dfrac {ca+b}{b+1}\geq\dfrac {18}

{a+b+c+3}$

let:$\dfrac {ab+c}{c+1}+\dfrac {bc+a}{a+1}+\dfrac {ca+b}{b+1}=p$

Using $AP \geq GP$ , then the smallest value of $p$ occurs when :

$\dfrac {ab+c}{c+1}=\dfrac {bc+a}{a+1}=\dfrac {ca+b}{b+1}=\dfrac {ab+bc+ca+a+b+c}

{a+b+c+3}=k$

$\therefore P\geq 3k\geq \dfrac {18}{a+b+c+3}$

(this will happen when $a=b=c=1$