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I need help or guidance in solving this inequality that I am battling for 3 days now. I have tried everything that comes to mind, but I am stuck. The inequality is as $$\sum_\textrm{cyc}\frac{1}{1+a+b^2}\leq1,\ abc=1, \ a,b,c>0$$ This can be written as $$\frac{1}{1+a+b^2}+\frac{1}{1+b+c^2}+\frac{1}{1+c+a^2}\leq1$$

I have tried Jensen's on the function $\frac{1}{1+x}$, AM-GM on the denominator, Cauchy-Schwarz for each term but I keep getting stuck.

kantorovich
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  • Can you please make clear what exactly you have to prove? – Qwerty Jun 15 '16 at 09:35
  • @SubhadeepDey And you are only given $abc=1$? – Qwerty Jun 15 '16 at 09:39
  • Yes @Qwerty with the condition that a,b,c are positive – kantorovich Jun 15 '16 at 09:41
  • seen this before and inequalities like this before. Will get back to ya soon. – DeepSea Jun 15 '16 at 10:00
  • sure, if there is a hint/direction in the meanwhile to keep me busy, it will be appreciated. ofcourse i will update here once any progress is made. – kantorovich Jun 15 '16 at 10:04
  • After making some computation, I got that the inequality is equivalent to $$a+b+c + a^2+b^2+c^2 \leq \frac{a^2}{c} + \frac{b^2}{a}+\frac{c^2}{b} + \frac{a}{b^2}+\frac{b}{c^2}+\frac{c}{a^2}$$. Hope this help! – GAVD Jun 15 '16 at 10:35
  • This inequality is quite sharp. The proof by @Qwerty is obviously not correct. I will post a nice proof in a few minutes. This type of inequalities appeared many times in math Olympiads, and there is a technique for it. – f10w Jun 15 '16 at 11:35

1 Answers1

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The following technique can deal with many of the above type of inequalities. For example, Problem 3 of IMO 2005 can be solved using the same idea (but a little bit easier). People participating in math Olympiads should keep it in mind.

For any $k$, applying Cauchy-Schwarz inequality we have $$(1+a+b^2)(c^{2k} + a^{2k-1} + b^{2k-2}) \ge (c^k+a^k+b^k)^2,$$ yielding $$\frac{1}{1+a+b^2} \le \frac{c^{2k} + a^{2k-1} + b^{2k-2}}{(a^k+b^k+c^k)^2}.$$ Similarly for the other two terms, taking the sum, it remains to prove $$\sum (a^{2k} + a^{2k-1} + a^{2k-2}) \le (a^k+b^k+c^k)^2$$ or equivalenty $$(a^{2k-1} + b^{2k-1} + c^{2k-1}) + (a^{2k-2} + b^{2k-2} + c^{2k-2}) \le 2(a^kb^k+b^kc^k+c^ka^k).$$ Re-writing this in homogeneous form: $$(abc)^{1/3}(a^{2k-1} + b^{2k-1} + c^{2k-1}) + (abc)^{2/3}(a^{2k-2} + b^{2k-2} + c^{2k-2}) \le 2(a^kb^k+b^kc^k+c^ka^k) \quad (*)$$ It suffices to find a value of $k$ for which this inequality holds for any $a,b,c > 0$. In this case, $k=2/3$ is such a value. Indeed, if $k=2/3$, denote $x=a^{1/3},y=b^{1/3},z=c^{1/3}$, $(*)$ becomes $$xyz(x+y+z) + x^2y^2z^2\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right) \le 2(x^2y^2+y^2z^2+z^2x^2),$$ or equivalently $$xyz(x+y+z) \le x^2y^2+y^2z^2+z^2x^2,$$ which is just $rs+st+tr \le r^2+s^2+t^2$ with $r=xy,s=yz,t=zx$.

Remark.

  1. The inequality $(*)$ is not true for the trivial value $k=1$. (This was indeed my first try.)
  2. I could have not mentioned $k$ at all, just implicitly set it to $2/3$, and put $x=a^{1/3},y=b^{1/3},z=c^{1/3}$ at the beginning of the proof. However, I think it's useful to show how I obtained the solution.
f10w
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