An Inequality Obtained by Transitivity // Transitiivisuudella saatu epäyhtälö

I thought that the NEWER magazines wouldn't bring back as many fond memories as the OLD magazines.

Here I found some sheets with some inequalities and an attempt at an axiomatic treatment of natural numbers...

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

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

ANSWER CiP

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

                           Solution CiP

          In the hypothesis  $abc\geqslant 1$  we will show the inequality :

$2\sum a+2\leqslant (\sum a)(\sum ab)-1 \tag{1}$

where  $\sum a=a+b+c\;\;,\;\;\sum ab =ab+bc+ca$.

From AM-GM Inequality  $x+y+z\geqslant 3\sqrt[3]{xyz}$  we get

$a+b+c\geqslant 3\sqrt[3]{abc}\;\;\overset{abc\geqslant 1}{\geqslant}\;3 \tag{2}$

$ab+bc+ca\geqslant \sqrt[3]{a^2b^2c^2}\underset{abc\geqslant 1}{\geqslant}3\tag{3}$

So  $(3)\Rightarrow\sum ab-2\leqslant 1 \underset{(2)}{\Rightarrow}(\sum a)(\sum ab-2)\geqslant 3\cdot 1\;\;\Leftrightarrow$

$\Leftrightarrow \;(\sum a)(\sum ab)-2\sum a \geqslant 3\;\Leftrightarrow\;2\sum a +2\leqslant (\sum a)(\sum ab)-1$

and  (1)  is proven. From  $abc\geqslant 1$  it follows by  TRANSITIVITY  from  (1)  that :

$2+2\sum a \leqslant (\sum a)(\sum ab)-abc \tag{4}$

     Let's now calculate the expression  $E:=\frac{1}{1+b+c}+\frac{1}{1+c+a}+\frac{1}{1+a+b}$.

$E=\frac{(1+c+a)(1+a+b)+(1+b+c)(1+a+b)+(1+b+c)(1+c+a)}{(1+b+c)(1+c+a)(1+a+b)}\;;$

$E=\frac{3+4\sum a+3\sum ab+\sum a^2}{1+2\sum a +3\sum ab+\sum a^2+\sum a^2b +\sum ab^2 +2abc}\;.$

Now using the equalities :

$\sum a^2=(\sum a)^2-2\sum ab$

$\sum a^2b+\sum ab^2=(\sum a)(\sum ab)-3abc$

then  $E$  takes its final form :

$E=\frac{3+4\sum a+\sum ab+(\sum a)^2}{1+2\sum a+\sum ab+(\sum a)^2+(\sum a)(\sum ab)-abc} \tag{5}$

We have  $E\leqslant 1$  if and only if  $2+2\sum a \leqslant (\sum a)(\sum ab)-abc$  what exactly is the inequality  (4).

     The  $"="$  sign occurs when we have equality in  (2), (3) and  $abc=1$  so  $a=b=c=1$.

$\blacksquare$

That's how I found it between the pages of the magazine

I have the feeling that the solution is incorrect...