Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that: $$\frac{1}{a+3}+\frac{1}{b+3}+\frac{1}{c+3}\geq\frac{a}{a^2+3}+\frac{b}{b^2+3}+\frac{c}{c^2+3}$$

I tried TL, BW, the Vasc's Theorems and more, but without success.

I proved this inequality!

I proved also the hardest version: $\sum\limits_{cyc}\frac{1}{a+4}\geq\sum\limits_{cyc}\frac{a}{a^2+4}$.

Thanks all!

  • 1
    $\begingroup$ I suppose if you want to limit your reader ship, abbreviations are a good plan, but ... $\endgroup$ – Thomas Andrews Jan 9 '17 at 20:04
  • $\begingroup$ i have a proof with BW Michael $\endgroup$ – Dr. Sonnhard Graubner Jan 9 '17 at 20:07
  • $\begingroup$ Sonnhard, I checked it again and I think BW does not help here. You are welcome to show us your proof and I'll find a mistake. $\endgroup$ – Michael Rozenberg Jan 9 '17 at 20:28
  • $\begingroup$ @Thomas Andrews I did not understand, what you said. $\endgroup$ – Michael Rozenberg Jan 9 '17 at 20:30
  • $\begingroup$ It means he doesn't know what TL and BW is. $\endgroup$ – Rutger Moody Jan 9 '17 at 21:45

BW in the following version does not help.

Let $a=x^3$, $b=y^3$ and $c=z^3$.

Hence, we need to prove that $$\sum_{cyc}\frac{1}{x^3+3xyz}\geq\sum_{cyc}\frac{x^3}{x^6+3x^2y^2z^2}$$ or $$\sum_{cyc}\frac{1}{x^3+3xyz}\geq\sum_{cyc}\frac{x}{x^4+3y^2z^2}.$$

Now, we can assume that $x=\min\{x,y,z\}$, $y=x+u$ and $z=x+v$

and these substitutions give inequality, which I don't know to prove.

But we can use another BW!

Let $a=\frac{y}{x}$, $b=\frac{z}{y}$ and $c=\frac{x}{z}$, where $x$, $y$ and $z$ are positives.

Hence, we need to prove that $$\sum_{cyc}\frac{x}{3x+y}\geq\sum_{cyc}\frac{xy}{3x^2+y^2}$$ or $$\sum_{cyc}\frac{x^3-x^2y}{(3x+y)(3x^2+y^2)}\geq0.$$ Now, let $x=\min\{x,y,z\}$, $y=x+u$ and $z=x+v$.

Hence, we need to prove that $$128(u^2-uv+v^2)x^7+16(16u^3+23u^2v-15uv^2+16v^3)x^6+$$ $$+32(8u^4+27u^3v+12u^2v^2-11uv^3+8v^4)x^5+$$ $$+4(32u^5+193u^4v+266u^3v^2-42u^2v^3-33uv^4+32v^5)x^4+$$ $$+2(8u^6+178u^5v+435u^4v^2+152u^3v^3-99u^2v^4+30uv^5+8v^6)x^3+$$ $$+uv(45u^5+375u^4v+291u^3v^2-83u^2v^3+57uv^4+3v^5)x^2+$$ $$+2u^2v^2(24u^4+66u^3v-18u^2v^2+13uv^3+3v^4)x+$$ $$+u^3v^3(18u^3-6u^2v+3uv^2+v^3)\geq0,$$ which is obvious.


  • $\begingroup$ Can you tell why the last inequality is obvious? If possible please expand it. +1 for the solution! $\endgroup$ – Random-generator May 15 '17 at 16:43
  • $\begingroup$ @Lohith-kumar Because $u\geq0$, $u\geq0$, $x>0$, $u^2-uv+v^2\geq0$,$16u^3+23u^2v-15uv^2+16v^3\geq0$... $\endgroup$ – Michael Rozenberg May 16 '17 at 2:46
  • $\begingroup$ Okay! But how to generally prove the non-negativity of, let's say, $24u^4+66u^3v−18u^2v^2+13uv^3+3v^4$? $\endgroup$ – Random-generator May 16 '17 at 6:30
  • 1
    $\begingroup$ @Lohith-kumar Because $24u^4+66u^3v-18u^2v^2+13uv^3+3v^4\geq9u^3v-18u^2v^2+9uv^3=9uv(u-v)^2\geq0$. $\endgroup$ – Michael Rozenberg May 16 '17 at 8:25
  • $\begingroup$ For this proof, you did not even need to do the clever subtraction to rid one of the terms of a minus sign on its highest power of $v$. For the same problem replacing 3 with 4, you did need that subtraction. $\endgroup$ – Mark Fischler Jun 27 '17 at 22:19

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.