PROOF WITHOUT WORDS

 ALINA , no matter how many nicknames  YOU  hide under, I want to talk to the REAL BEING that you are. Otherwise, words have no meaning, like in this post.

          Here we present a Proof, "in one piece", of the Exercise from the posts of the past few days. We do it without words...

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

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

I have solved this problem correctly here. Now we put a wordless solution, not really in the spirit of the work

NELSEN Roger B.

Proofs Without Words : Exercises in Visual Thinking

The Mathematical Association of America, f.l., 1993

                              Proof CiP

$k^3=abc \geqslant 1 \tag{2}$

$a':=\frac{a}{k}\;,\;\;b':=\frac{b}{k}\;,\;\;c':=\frac{c}{k}\;\;\overset{k\geqslant 1}{\Rightarrow}$

$a'\leqslant a\;,\;\;b'\leqslant b\;,\;\;c'\leqslant c\;\;,\;\;\;a'\cdot b'\cdot c'\overset{(2)}{=}1 \tag{3}$

$a'=:x^3\;,\;\;b'=:y^3\;,\;\;c'=:z^3\;\;\;(x,\;y,\;z>0)\;\;\Rightarrow\;x\cdot y\cdot z=1 \tag{4}$

$y^3+z^3\geqslant y^2z+yz^2\;\;,\;\;"="\Leftrightarrow y=z \tag{5}$

[$(y-z)^2(y+z)\geqslant 0\;,\;\;"="\Leftrightarrow y=z\;\Rightarrow(y-z)(y^2-z^2)\geqslant 0\Rightarrow y^3-y^2z-yz^2+z^3\geqslant 0\Rightarrow (5)]$

$1+b+c\underset{(3)}{\geqslant} 1+b'+c'\underset{(4)}{=}xyz+y^3+z^3\underset{(5)}{\geqslant} xyz+y^2z+yz^2=yz(x+y+z) \tag{6}$

$\frac{1}{1+b+c}\overset{(6)}{\underset{(4)}{\leqslant}} \frac{xyz}{yz(x+y+z}=\frac{x}{x+y+z}\;\;\;(7)\;\Rightarrow$

$$\sum_{cycl}\frac{1}{1+b+c}\overset{(7)}{\leqslant}\sum_{cycl}\frac{x}{x+y+z}=\frac{x+y+z}{x+y+z}=1.$$

$"="\;in\;(1)\Leftrightarrow "="\;in\;(7)\;(6)\;(5)\;(3)\;(2)\;\;\Leftrightarrow x=y=z\;,\;k=1\;\Leftrightarrow a=b=c=1$

$\blacksquare\;\;\;\;\blacksquare$