Algebra question : Prove the inequality.

Question

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 : Equation

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 :

Pretty-Equation

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

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

Pretty -3

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 09:48

score 3 · Answer 1 · edited Jun 30 '14 at 12:12

3

enter image description here Using the given inequality \begin{align*} & a+b+c-(ab+bc+ca)\ge 0\implies \dfrac{a+b+c}{ab+bc+ca}\ge 1\\ & (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\\ \implies & \dfrac{(a+b+c)^2}{ab+bc+ca}=\dfrac{a^2+b^2+c^2}{ab+bc+ca}+2\\ \implies & (a+b+c)\left(\dfrac{a+b+c}{ab+bc+ca}+1\right)=\dfrac{a^2+b^2+c^2}{ab+bc+ca} +(a+b+c)+2\\ \implies & 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})≤2(a+b+c)≤\dfrac{a^2+b^2+c^2}{ab+bc+ca}+(a+b+c)+2 \end{align*}

edited Jun 30 '14 at 12:12

shadow10

5,616

answered Jun 30 '14 at 11:00

Bumblebee

18,220
5
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The first inequality can be upgraded to strict inequalities, leading to the final inequality upgradable also. Guess the non-strict inequality in the problem statement is rather misleading. – Gina Jun 30 '14 at 17:37
Can anyone also comment on the methods I tried? Because I think there may be alter methods too for this question. – Kushashwa Ravi Shrimali Jul 01 '14 at 00:25
@Nilan.C.Manoj : How did you get to $$ a + b + c -(ab + bc + ca) \ge 0 $$ – Kushashwa Ravi Shrimali Jul 01 '14 at 09:35
For any non negative a and b, we have a≥a/(1+b) – Bumblebee Jul 01 '14 at 09:41
Oh, I see. Is there any proof for this too? – Kushashwa Ravi Shrimali Jul 01 '14 at 09:43
And Nilan, can you please type the whole thing for how you got to a + b + c - (ab + bc + ca) >= 0 . As, I don't know how the above inequality you typed is connected to the quest. – Kushashwa Ravi Shrimali Jul 01 '14 at 09:49
I think it is obvious. Am I done anything wrong? – Bumblebee Jul 01 '14 at 09:50
Try to replace a by a-bc and b by b+c. I think it will work. – Bumblebee Jul 01 '14 at 09:52
I am not sure whether you have done anything wrong or not. I'm just thinking that how exactly you got to the inequality. $$a + b + c - (ab + bc + ca) \ge 0 $$ – Kushashwa Ravi Shrimali Jul 01 '14 at 09:52
Let us continue this discussion in chat. – Kushashwa Ravi Shrimali Jul 01 '14 at 09:53
Please come there. @NilanC.Manoj – Kushashwa Ravi Shrimali Jul 01 '14 at 09:54
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@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 02 '14 at 16:18
@Gina I will give it a try. Thanks! And here is what I just got from a user (who created this problem) : http://emc.mnm.hr/wp-content/uploads/2013/12/EMC-2013-Junior-Category-Solutions.pdf – Kushashwa Ravi Shrimali Jul 02 '14 at 16:55
@Gina,it is not true. you don't know $a-bc>0$. – chenbai Jul 04 '14 at 00:39
this proof is not correct. $a=2,b=c=0.8$ is a counter example. – chenbai Jul 04 '14 at 01:01
@chenbai: ah thanks for pointing that out, I missed that. I guess the more complicated solution is the only one we got then. – Gina Jul 04 '14 at 01:07

Algebra question : Prove the inequality.

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