# For $abc=1$ prove that $\sum\limits_\text{cyc}\frac{1}{a+3}\geq\sum\limits_\text{cyc}\frac{a}{a^2+3}$

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!

• I suppose if you want to limit your reader ship, abbreviations are a good plan, but ... – Thomas Andrews Jan 9 '17 at 20:04
• i have a proof with BW Michael – Dr. Sonnhard Graubner Jan 9 '17 at 20:07
• 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. – Michael Rozenberg Jan 9 '17 at 20:28
• @Thomas Andrews I did not understand, what you said. – Michael Rozenberg Jan 9 '17 at 20:30
• It means he doesn't know what TL and BW is. – 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.

Done!

• Can you tell why the last inequality is obvious? If possible please expand it. +1 for the solution! – Random-generator May 15 '17 at 16:43
• @Lohith-kumar Because $u\geq0$, $u\geq0$, $x>0$, $u^2-uv+v^2\geq0$,$16u^3+23u^2v-15uv^2+16v^3\geq0$... – Michael Rozenberg May 16 '17 at 2:46
• Okay! But how to generally prove the non-negativity of, let's say, $24u^4+66u^3v−18u^2v^2+13uv^3+3v^4$? – Random-generator May 16 '17 at 6:30
• @Lohith-kumar Because $24u^4+66u^3v-18u^2v^2+13uv^3+3v^4\geq9u^3v-18u^2v^2+9uv^3=9uv(u-v)^2\geq0$. – Michael Rozenberg May 16 '17 at 8:25
• 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. – Mark Fischler Jun 27 '17 at 22:19