Prove Inequality: $ab+c, bc+a, ca+b \geq 18$

[Albert1]

$a,b,c\geq 1$

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

{a+b+c+3}$

 

[lfdahl]

Spoiler

\[\frac{ab+c}{c+1}+\frac{bc+a}{a+1}+\frac{ca+b}{b+1}\geq \frac{18}{(a+1)+(b+1)+(c+1)}\\ \frac{1}{2}(ab+c + bc+a+ca+b)\geq \frac{18}{(a+1)+(b+1)+(c+1)}\\ c(a+1)+a(b+1)+b(c+1)\geq \frac{36}{(a+1)+(b+1)+(c+1)}\\ (a+1)+(b+1)+(c+1)\geq \frac{36}{(a+1)+(b+1)+(c+1)}\\ \left [ (a+1)+(b+1)+(c+1)\right ]^2\geq 36\\\]

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

 

[Albert1]

$a,b,c\geq 1$

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

{a+b+c+3}$

Spoiler

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 $a=b=c=1$