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

Question

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 09 '17 at 20:04
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 09 '17 at 20:28
@Rutger Moody TL it's Tangent Line method. See here http://math.stackexchange.com/questions/2081728/given-that-xyz-1-prove-that-fracx1x4-fracy1y4-fracz1z4 an example how it works. About BW see here: http://www.artofproblemsolving.com/community/c6h522084 — Michael Rozenberg, Jan 09 '17 at 21:53
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! — Michael Rozenberg, Mar 31 '17 at 04:29
You mentioned that you used Vasc's theorems. I know of Vasc's Inequality, but what are Vasc's theorems, and where are they? Do you mean his methods like $n-1$ EV, etc? — Book Of Flames, Oct 31 '20 at 08:24
@Book Of Flames I meant Vasc's RCF Theorem, LCF Theorem, HCF Theorem, E-V Method — Michael Rozenberg, Oct 31 '20 at 08:26

score 3 · Answer 1 · answered Mar 31 '17 at 05:26

3

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!

answered Mar 31 '17 at 05:26

Michael Rozenberg

194,933

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 02:46
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 06:30
1

@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 08:25
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

score 1 · Answer 2 · answered Dec 22 '22 at 17:06

Another way.

Since for any $a>0$ we have $$\frac{1}{a+3}-\frac{a}{a^2+3}+\frac{9}{64}\geq\frac{27}{64\left(a^{\frac{8}{3}}+a^{\frac{4}{3}}+1\right)}$$ and for positives $a$, $b$ and $c$ such that $abc=1$ we have $$\sum_{cyc}\frac{1}{a^2+a+1}\geq1,$$ we obtain: $$\sum_{cyc}\left(\frac{1}{a+3}-\frac{a}{a^2+3}\right)=\sum_{cyc}\left(\frac{1}{a+3}-\frac{a}{a^2+3}+\frac{9}{64}\right)-\frac{27}{64}\geq$$ $$\geq\frac{27}{64}\left(\sum_{cyc}\frac{1}{a^{\frac{8}{3}}+a^{\frac{4}{3}}+1}-1\right)\geq0.$$

Miss and Mister cassoulet char · Answer 3 · 2022-12-24T10:24:09.923

0

Partial Hint and too long for a comment :

Case where two varaibles are superior two one .

If we define :

$$f\left(x\right)=\frac{1}{e^{x}+3}-\frac{e^{x}}{e^{2x}+3},g\left(x\right)=\ln\left(1+e^{-x}\right)-\ln\left(2\right)$$

Show that for $x\in(-\infty,\infty)$ :

$$f''(x)+g''(x)>0$$

Now apply Jensen's inequality on two variable .

Remains to show :

$$-\left(g\left(a\right)+g\left(b\right)-2g\left(\frac{-c}{2}\right)\right)+f\left(c\right)\geq 0$$

Or :

$$-\left(g\left(a\right)+g\left(b\right)-2g\left(\frac{-c}{2}\right)\right)+\ln\left(f\left(c\right)+1\right)\geq 0$$

Wich is true for $a+b+c=0, a,b\in(0,2) \operatorname{or} a,b\in(2,\infty)$,

Last hint remark with the constraint $x,y\in [1,\infty)$:

$$\frac{\left(1+\frac{1}{xy+3}-\frac{xy}{\left(xy\right)^{2}+3}\right)\left(1+\sqrt{xy}\right)^{2}}{\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)}\ge \frac{\left(3+\frac{1}{xy+3}-\frac{xy}{\left(xy\right)^{2}+3}\right)\left(3+\sqrt{xy}\right)^{2}}{3\left(3+\frac{1}{x}\right)\left(3+\frac{1}{y}\right)}\ge \frac{4\left(3+\frac{1}{xy+3}-\frac{xy}{\left(xy\right)^{2}+3}\right)\left(3+\sqrt{xy}\right)}{3\left(3+\frac{1}{x}\right)\left(3+\frac{1}{y}\right)}\ge1$$

Case where two variable are less than one

Now consider the function :

$$p\left(x\right)=\frac{1}{x+3}-\frac{x}{x^{2}+3}$$

There exists a positive constant $C$ and $b\in(0,1),x>0$ such that :

$$p\left(Cx\right)-p\left(Cx^{b}\right)\geq 0$$

$$C=\left(3-2\sqrt{2}\right)^{\frac{1}{3}}+\left(3+2\sqrt{2}\right)^{\frac{1}{3}}$$

To be continued...

edited Dec 24 '22 at 10:24

answered Dec 22 '22 at 13:22

Miss and Mister cassoulet char

3,511

Why your reasoning helps? – Michael Rozenberg Dec 22 '22 at 13:27
@MichaelRozenberg Hi, I have the feeling you want details...But I haven't the time now and I don't forget you . – Miss and Mister cassoulet char Dec 22 '22 at 13:54
@ErikSatie It does not help. You only proved $f(x) + f(y) + f(z) \ge 0$ for $x + y + z = 0$. However, this does not imply $\sum_{\mathrm{cyc}} [\frac{1}{e^{x}+3}-\frac{e^{x}}{e^{2x}+3}] \ge 0$ for $x + y + z = 0$ since $\sum_{\mathrm{cyc}} [\ln\left(1+e^{-x}\right)-\ln\left(2\right)] \ge 0$ for $x + y + z = 0$. – River Li Dec 23 '22 at 02:45
@RiverLi Yes I agree . – Miss and Mister cassoulet char Dec 23 '22 at 10:11
@RiverLi I propose now a partial hint . – Miss and Mister cassoulet char Dec 23 '22 at 10:50
@RiverLi I have alsmot a proof what do you think about ?Thanks and happy merry christmas! – Miss and Mister cassoulet char Dec 23 '22 at 14:23
@ErikSatie Happy Merry Christmas! (Your answer seems not helpful.) – River Li Dec 23 '22 at 14:38
For someone who want to show the last assertions consider $x^y=1+a+b,x=a+b,a,b\in(0,\infty)$ all the interesting coefficients are negatives .We can do the same thing with $x\in(0,1)$ wich I let to the reader as an exercise. – Miss and Mister cassoulet char Dec 24 '22 at 13:51

score 0 · Answer 4 · answered Dec 23 '22 at 02:36

Alternative proof:

WLOG, assume that $a \le b \le c$.

We split into two cases:

Case 1: $c \le 8/5$

We have, for all $0 < x \le 8/5$, $$\frac{1}{x + 3} - \frac{x}{x^2 + 3} - \frac{28x^2 - 101x + 73}{240} = \frac{(-28x^3 - 39x^2 + 96x + 63)(x - 1)^2}{240(x+3)(x^2+3)} \ge 0.$$

It suffices to prove that $$\sum_{\mathrm{cyc}} \frac{28a^2 - 101a + 73}{240} \ge 0$$ or $$28(a^2 + b^2 + c^2) - 101(a + b + c) + 219 \ge 0. \tag{1}$$ (Note: Actually, (1) is true for all $a, b, c\ge 0$ with $abc = 1$.)

The proof of (1) using pqr method is given at the end.

$\phantom{2}$

Case 2: $c > 8/5$

Let $f(x) = \frac{1}{x + 3} - \frac{x}{x^2 + 3}$.

We have, for all $x > 0$, $$f(x) + \frac{8}{90} = \frac{4x^3 + 12x^2 - 123x + 171}{45(x+3)(x^2+3)} \ge 0. \tag{2}$$

Note that $f(x)$ is strictly decreasing on $(0, 2)$.

If $a \le 5/16$, using (2), we have $$f(a) + f(b) + f(c) \ge f(5/16) - \frac{8}{90} - \frac{8}{90} > 0.$$

If $a > 5/16$, we have $b = \frac{1}{ca} \le \frac{5}{8a} < 2$ and thus (using (2)) $$f(a) + f(b) + f(c) \ge f(a) + f(\tfrac{5}{8a}) - \frac{8}{90} > 0.$$

We are done.

Proof of (1):

We use the pqr method.

Let $p = a + b + c, q = ab + bc + ca, r = abc = 1$.

We need to prove that $28p^2 - 56q - 101p + 219 \ge 0$.

Degree 4 Schur yields $$4q^2 - 5p^2q + p^4 + 6pr \ge 0$$ which results in (using $p^2 \ge 3q$) $$q \le \frac58 p^2 - \frac18\sqrt{9p^4 - 96pr } = \frac58 p^2 - \frac18\sqrt{9p^4 - 96p }.$$ We have \begin{align*} &28p^2 - 56q - 101p + 219\\ \ge\,& 28p^2 - 56\left(\frac58 p^2 - \frac18\sqrt{9p^4 - 96p }\right) - 101p + 219\\ =\,& 7\sqrt{9p^4 - 96p} - (7p^2 + 101p - 219)\\ \ge\,& 0 \end{align*} where we have used $p \ge 3$.

We are done.

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

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