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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!

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  • $\begingroup$ Gerry Myerson, thanks for editing the tags. Can you please help me in this question? $\endgroup$ – Kushashwa Ravi Shrimali Jun 30 '14 at 9:48
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enter image description hereUsing 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*}

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  • $\begingroup$ 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. $\endgroup$ – Gina Jun 30 '14 at 17:37
  • $\begingroup$ Can anyone also comment on the methods I tried? Because I think there may be alter methods too for this question. $\endgroup$ – Kushashwa Ravi Shrimali Jul 1 '14 at 0:25
  • $\begingroup$ @Nilan.C.Manoj : How did you get to $$ a + b + c -(ab + bc + ca) \ge 0 $$ $\endgroup$ – Kushashwa Ravi Shrimali Jul 1 '14 at 9:35
  • $\begingroup$ For any non negative a and b, we have a≥a/(1+b) $\endgroup$ – Bumblebee Jul 1 '14 at 9:41
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    $\begingroup$ @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 $a-bc+b-ca+c-ab>0$. $\endgroup$ – Gina Jul 2 '14 at 16:18

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