CORRECTION of the previous "PROOF"

Since all communication with ALINA has been broken,

 we spend our days and nights talking nonsense.

 But my thoughts are always on HER.                    

 A COMMENTATOR, unfortunately anonymous, drew my attention to the fact that the solution is WRONG. I tried to explain it to him, but in truth, I had misapplied the most elementary inequality techniques.

COPILOT saved me this time too, see his draft at the end. Now I'm just writing the CORECT Solution.

          Let's resume the statement of the Problem.

          "Let  $a,\;b,\;c$  be real positive numbers such that  $abc\geqslant 1$. Show that

$\frac{1}{1+b+c}+\frac{1}{1+c+a}+\frac{1}{1+a+b}\leqslant 1" \tag{1}$

ANSWER CiP

$"="\;\Leftrightarrow\;a=b=c=1$

                 Solution CiP  with COPILOT assistant

               We first consider the CASE  $\fbox{$abc=1$}$

          Let  $x,\;y,\;zy,z\;>0$  be such that  $a=x^3,\;b=y^3,\;c=z^3$. In this case we have

$xyz=1 \tag{2}$

We notice that, because  $y^3-y^2z-yz^2+z^3=(y-z)^2(y+z)\geqslant 0\;\;with\;"="\;iff\;y=z$ , it takes place

$y^3+z^3\geqslant y^2z+yz^2 \tag{3}$

We write

$\frac{1}{1+b+c}\overset{(2)}{=}\frac{1}{xyz+y^3+z^3}\overset{(3)}{\leqslant} \frac{1}{xyz+y^2z+yz^2}=\frac{1}{yz(x+y+z)}\overset{(2)}{=}\frac{x}{x+y+z}$

Writing the other two analogous relations, and adding them, we get

$\sum \frac{1}{1+b+c}\leqslant \sum \frac{x}{x+y+z}=\frac{x+y+z}{x+y+z}=1$

     So the inequality is proven in this case; "="  occurs if and only if  $x=y=z$  , so  $a=b=c=1$.

               $\fbox{$abc>1$}$

          Let  $abc=k^3\;,\;a'=\frac{a}{k},\;b'=\frac{b}{k},\;c'=\frac{c}{k}.$ Therefore

$a'b'c'=1\;\;;\;\;k>1 \tag{4}$

But  $1+b'+c'=1+\frac{b}{k}+\frac{c}{k}\overset{k>1}{<}1+b+c$  so

$\frac{1}{1+b+c}<\frac{1}{1+b'+c'}$

Then  $\sum\frac{1}{1+b+c}<\sum\frac{1}{1+b'+c'}\leqslant 1$  becase for  $a',\;b',\;c'$  we fall over the previous case.

     We note that in the second case the sign  "="  cannot occur.

$\blacksquare$

Draft of the conversation with COPILOT, who came up with this solution idea :

Ideea e foarte simpatică: mai întâi rezolvăm cazul $abc=1$, apoi îl reducem pe cel cu $abc\ge 1$ la acesta.

1. Cazul $abc=1$

Presupunem $a,b,c>0$ și $abc=1$. Scriem

a=x3,b=y3,c=z3,xyz=1.

Atunci

11+b+c=11+y3+z3,etc.

Observăm inegalitatea

$$x^3+y^3\ge x^{2}y+y^{2}x,$$

pentru că

$$x^3+y^3-x^{2}y-y^{2}x=x^2(x-y)+y^2(y-x)=(x-y{)}^2(x+y)\ge 0.$$

Deci

$$\frac{1}{x^3+y^3+xyz}\le \frac{1}{x^{2}y+y^{2}x+xyz}\mathrm{.}$$

Aplicând ciclic,

$$\sum _{\text{ciclic}}\frac{1}{a+b+1}=\sum _{\text{ciclic}}\frac{1}{x^3+y^3+xyz}\le \sum _{\text{ciclic}}\frac{1}{x^{2}y+y^{2}x+xyz}\mathrm{.}$$

Dar

$$x^{2}y+y^{2}x+xyz=xy(x+y+z),$$

deci

$$\frac{1}{x^{2}y+y^{2}x+xyz}=\frac{1}{xy(x+y+z)}=\frac{z}{x+y+z},$$

folosind $xyz=1$. Prin ciclicitate:

$$\sum _{\text{ciclic}}\frac{1}{x^{2}y+y^{2}x+xyz}=\sum _{\text{ciclic}}\frac{z}{x+y+z}=\frac{x+y+z}{x+y+z}=1.$$

Rezultă

11+a+b+11+b+c+11+c+a≤1pentru abc=1.

2. Extinderea la $abc\ge 1$

Fie acum $abc\ge 1$. Notăm

k=(abc)1/3≥1,a′=ak, b′=bk, c′=ck.

Atunci

$$a^{\mathrm{\prime }}{b}^{\mathrm{\prime }}{c}^{\mathrm{\prime }}=\frac{abc}{k^3}=1,$$

deci pentru $a^{\mathrm{\prime }},b^{\mathrm{\prime }},c^{\mathrm{\prime }}$ avem deja

$$\frac{1}{1+a^{\mathrm{\prime }}+b^{\mathrm{\prime }}}+\frac{1}{1+b^{\mathrm{\prime }}+c^{\mathrm{\prime }}}+\frac{1}{1+c^{\mathrm{\prime }}+a^{\mathrm{\prime }}}\le 1.$$

În plus, cum $k\ge 1$,

$$1+a^{\mathrm{\prime }}+b^{\mathrm{\prime }}=1+\frac{a}{k}+\frac{b}{k}\le 1+a+b,$$

de unde

$$\frac{1}{1+a+b}\le \frac{1}{1+a^{\mathrm{\prime }}+b^{\mathrm{\prime }}}\mathrm{.}$$

La fel pentru celelalte două fracții.

Prin urmare,

$$\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\le \frac{1}{1+a^{\mathrm{\prime }}+b^{\mathrm{\prime }}}+\frac{1}{1+b^{\mathrm{\prime }}+c^{\mathrm{\prime }}}+\frac{1}{1+c^{\mathrm{\prime }}+a^{\mathrm{\prime }}}\le 1.$$

Și gata: pentru orice $a,b,c>0$ cu $abc\ge 1$,

$$\boxed{{\displaystyle \frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a}\le 1}}$$

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