Just as I am not allowed to forget HER
The author of the problem, Vasile BIVOLARU, was my high school teacher. I am posting it here in his memory.
"24063. (page 85) Prove that if $a>1$ , then for any $x\in\mathbb{R}$ ,
the following inequality holds:
$a^x+(a+1)^x+(a^2+5a+6)^x\geqslant (a+2)^x+(a+3)^x+(a^2+a)^x$
{Author : } Vasile BIVOLARU, Timișoara"
ANSWER CiP
$"="\;\;\Leftrightarrow\;\;x=0$
Solution CiP
Let us denote :
$b=a^x\;\;,\;\;c=(a+1)^x\;\;,\;\;u=(a+2)^x\;\;,\;\;v=(a+3)^x$
Noting that $a^2+5a+6=(a+2)(a+3)\;\;and\;\;a^2+a=a(a+1)$ , the inequality to be proven is written as follows:
$b+c+u\cdot v \geqslant u+v+b\cdot c\;\;\Leftrightarrow$
$\Leftrightarrow\;\;u\cdot v-u-v+1\geqslant b\cdot c-b-c+1\;\;\Leftrightarrow$
$\Leftrightarrow\;\;(u-1)(v-1)\geqslant (b-1)(c-1) \tag{1}$
If $x>0$ , then $v>u>c>b>1$ so
$\begin{cases}u-1>b-1>0\\v-1>c-1>0\end{cases} \tag{2}$
and multiplying the inequalities (2) above, we obtain (1) , with the sign $">"$.
If $x<0$ , then $1>b>c>u>v>0$ so $1-v>1-u>1-c>1-b>0$ , and then
$(1-v)(1-u)>(1-c)(1-b)\;\;\Leftrightarrow\;\;(v-1)(u-1)>(c-1)(b-1)$
and we arrive at the same conclusion.
If $x=0$ then $b=c=u=v=1$ and we obtain $"="$ in (1).
We obtained the answer.
$\blacksquare$
REMARK CiP I would not have easily found this beautiful and simple solution without the help of AOPS.
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