Prove that, if $a,b,c \geq 1$, then $\frac{3ab+2c+1}{a+b}+\frac{3bc+2a+1}{b+c}+\frac{3ca+2b+1}{c+a} \geq 9$.

Question

the question

Prove that, if $a,b,c \geq 1$, then

$\frac{3ab+2c+1}{a+b}+\frac{3bc+2a+1}{b+c}+\frac{3ca+2b+1}{c+a} \geq 9$.

the idea

I know that these arent really helpful ideas, but as you know we have to use an inequality theorem to demonstrate this....I thought of the inequality of means, but when I tried the equality case it didn't work, so we I can't show the inequality by the inequality of means

Then I thought of Bermstrong or CBS and tried writing LHS in different forms that are probably useful.

I always get stuck at this kind of problem, when I get to show inequality. Hope one of you can help me! Thank you!

Answer 1

Use the fact that $$3ab+1\ge 2ab+2\ge 2a+2b \hspace{1cm} \text{for }a,b\ge 1$$ Then use C-S inequality $$\sum_{\text{sym}}\frac{3ab+2c+1}{a+b} \ge\sum_{\text{sym}}\frac{2a + 2b +2c}{a+b} = 2(a+b+c) \sum_{\text{sym}}\frac{1}{a+b} \ge 9$$ The equality occurs if and only if $a = b=c =1$.