# Algebra question : Prove the inequality.

Let $a , b \ \& \ c$ be positive real numbers satisfying : $$\cfrac{a}{1+b+c} + \cfrac{b}{1+c+a} + \cfrac{c}{1+a+b} \ge \cfrac{ab}{1+a+b} + \cfrac{bc}{1+b+c}+ \cfrac{ca}{1+a+c}$$ Prove that : $$\cfrac{a^2 + b^2 + c^2}{ab + bc + ca} + a + b + c + 2 \ge 2 \left(\sqrt{ab} + \sqrt{bc} + \sqrt{bc} \right)$$

This is what I've tried yet :

While, I know that this is not anything like appreciable effort. But, I really have tried a lot after this including the logic of all the denominators being positive.

This is the try for the first equation which didn't lead me anywhere. I tried to simplify the second equation also, this is what I got :

This looked very good to me. But, again my brain is empty with further ideas.

The ideas I have got till now include : $$\bullet \text{Using AM-GM inequality} :$$

I don't know what to do further. I thought about this for 1 day and then at each break, an idea came in my mind but all were in vain.

I will appreciate any help regarding this. And if anyone could comment on my ideas and try to explain how to go on from there too (if there is any way) , then it will be great also.

Thanks!

• Gerry Myerson, thanks for editing the tags. Can you please help me in this question? – Kushashwa Ravi Shrimali Jun 30 '14 at 9:48

• @Nilan.C.Manoj : How did you get to $$a + b + c -(ab + bc + ca) \ge 0$$ – Kushashwa Ravi Shrimali Jul 1 '14 at 9:35
• @KushashwaRaviShrimali: move the right side to the left side, you end up with one term that look like $\frac{a-bc}{1+b+c}$. Then $a-bc>\frac{a-bc}{1+b+c}$ since $1+b+c>1$. Do that for all 3 term. Then you will get $a-bc+b-ca+c-ab>0$. – Gina Jul 2 '14 at 16:18