Au trecut două zile de când nu a mai fost Alina pe chat... Îmi amintesc că odată, a intrat cu nickul "lookigforsummer"... Am vorbit cu EA ulterior si mi-a spus ca este titlul unei melodii, Looking For The Summer ...
Am mai văzut din când în când un nick, care am crezut ca este al ei : "weakmeupwhenseptemberends"... Sigur i-ar fi placut... Dar nu, am vorbit cu respectivul... Este tot titlul unei melodii...
Today I scanned another magazine, from my increasingly incomplete collection... Older issues are here.
"E:12378. If $a\in (0\;,\;8)$ , prove that
$\left ( 1+\frac{1}{\sqrt{a}} \right )\cdot \left (1+\frac{1}{\sqrt{8-a}}\right )\geqslant \frac{9}{4}.$
. {Author :} Gh ACHIM, Mizil, Prahova"
ANSWER CiP
"=" iff a=4
Solution CiP
We have the inequality :
$x+y\geq 2\sqrt{xy}\;\;,\;\;x,y\geqslant 0 \;\;,\;\;"="\Leftrightarrow x=y \tag{I}$
Indeed, $(\sqrt{x}-\sqrt{y})^2\geqslant 0\;\;"=" \Leftrightarrow x=y$. So $x+y-2\sqrt{xy}\geqslant 0$...etc
Let $b:=\sqrt{a}\;\;,\;\;c:=\sqrt{8-a}$ ; we have
$b^2+c^2=8 \tag{1}$
Writing (I) for $x=b^2\;,\;y=c^2$ we get $b^2+c^2\geqslant 2bc\;\;\overset{(1)}{\Rightarrow} 8\geqslant 2bc$ so
$\frac{1}{bc}\geqslant \frac{1}{4} \tag{2}$
Writing (I) for $x=\frac{1}{b}\;,\;y=\frac{1}{c}$ we get $\frac{1}{b}+\frac{1}{c}\geqslant \frac{2}{\sqrt{bc}}\underset{(2)}{\geqslant} \frac{2}{\sqrt{4}}$ so
$\frac{1}{b}+\frac{1}{c}\geqslant 1 \tag{3}$
Then $1+\left (\frac{1}{b}+\frac{1}{c}\right )+\frac{1}{bc}\overset{(2)\;(3)}{\geqslant} 1+1+\frac{1}{4}\Leftrightarrow$
$\Leftrightarrow \left (1+\frac{1}{b}\right )\cdot \left (1+\frac{1}{c}\right )\geqslant \frac{9}{4}$
that is, the inequality to be proved.
$"="\Leftrightarrow b=c\;and\;\frac{1}{b}=\frac{1}{c}\;\;\Leftrightarrow b=c=\sqrt{\frac{b^2+c^2}{2}}\underset{(1)}{=}2\;\Leftrightarrow a=4.$
$\blacksquare$
