Let $a,b,c,d>0$ and $a+b+c+d=1$. Prove that $\dfrac{abc}{1+bc}+\dfrac{bcd}{1+cd}+\dfrac{cda}{1+ad}+\dfrac{dab}{1+ab}\le \dfrac{1}{17}$

My attempt:

I figured out that if each of the element could be like $\dfrac{abc}{1+bc}\le \dfrac{1}{68}$ then we would be done.

From a little manipulation we get, $\dfrac{1}{bc}+1\le 68a$ , $\dfrac{1}{cd}+1\le 68b$ , $\dfrac{1}{da}+1\le 68c$ , $\dfrac{1}{ab}+1\le 68d$. Summing then we get,

$\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{cd}+\dfrac{1}{da}\le 64$

But, I suppose my first assumption could be wrong, if not then please help me further and if so then please help with the solution. Thank you.

  • $\begingroup$ @Macavity sorry, please see the edit $\endgroup$ – Swadhin Feb 5 '15 at 7:44
  • $\begingroup$ OK. Still $\dfrac{abc}{1+bc} \le \dfrac1{68}$ need not be true. There are counter examples, so that approach is doomed. $\endgroup$ – Macavity Feb 5 '15 at 7:45
  • $\begingroup$ Yes, I thought so, but I cannot approach the problem in any way. $\endgroup$ – Swadhin Feb 5 '15 at 7:49

You could approach it as follows: $$ \sum_{cyc} \frac{abc}{1+bc}=\sum_{cyc} a\left(1-\frac{1}{1+bc}\right)=\sum_{cyc} \left(a-\frac{a}{1+bc}\right)=1-\sum_{cyc} \frac{a}{1+bc} $$ So the inequality is equivalent to: $$ 1-\sum_{cyc} \frac{a}{1+bc}\le\frac{1}{17}\iff\frac{16}{17}\le\sum_{cyc} \frac{a}{1+bc} $$ Using CS, this can be reduced to prove the following: $$ \left(\sum_{cyc} \frac{a}{1+bc}\right)\cdot\left(\sum_{cyc} a(1+bc)\right)\ge(a+b+c+d)^2=1\iff\sum_{cyc} \frac{a}{1+bc}\ge\frac{1}{\sum_{cyc} a(1+bc)}=\frac{1}{a+b+c+d+abc+bcd+cda+dab}=\frac{1}{1+abc+bcd+cda+dab} $$ So if $$ \frac{1}{1+abc+bcd+cda+dab}\ge\frac{16}{17}\iff abc+bcd+cda+dab\le\frac{1}{16} $$ is true, the original inequality would be true as well.


The inequality $$ abc+bcd+cda+dab\le\frac{1}{16} $$ is true due to Maclaurin's inequality , which, in a special case, states that: $$ \left(\frac{abc+bcd+cda+dab}{4}\right)^{\frac13}\le\frac{a+b+c+d}{4}=\frac{1}{4}\iff abc+bcd+cda+dab\le\frac{1}{16} $$ And your inequality is proven.


Here is my approach. I am stuck at a place but I hope this is a good method and I don't reach a dead end. The LHS is equivalent to $\frac{a}{1/bc+1}+\frac{b}{1/cd+1}+\frac{c}{1/da+1}+\frac{d}{1/ab+1}$. From AM-GM, we have $1+1/bc>2/\sqrt{bc}$. Inverse the inequality with inverting the inequality sign and then multiply by $a$ and write the other 3 terms. Then add all the terms. Denote our original expression as P. We have $P<a\sqrt{bc}+b\sqrt{cd}+c\sqrt{da}+d\sqrt{ab}/2$. Which itself is less than $ab+bd+ab+ac+bc+bd+cd+ac/4$. Now I am working to prove this less than 1/17, hope this helps.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.