A Problem in Litigation : by Cristian OLTEANU

          The author of Problem #4 in the image is a diligent GMB contributor. The problem was given to 8th graders during the "Nicolae POPESCU" Mathematics Competition...

            With the data in the figure, the problem has no solution. I modified it to my taste, obtaining:

           4.    Calculate the minimum of the expression $\frac{1}{ab}+9ab$, knowing that

                  $(3a+2)(b+1)=6,\;a,b \in (0,+\infty)$, and find the values ​​of the

                  numbers $a$ and $b$ for which this minimum is obtained.

ANSWER CiP

$$\frac{1}{ab}+9ab\geqslant 6;\;\;Equality\;for\;a=\frac{1}{3},\;b=1\;\;or\;\;a=\frac{2}{3},\;b=\frac{1}{2}$$

                     Solution CiP

             $\frac{1}{ab}+9ab=3 \cdot \left( \frac{1}{3ab}+3ab\right )\geqslant 3 \cdot 2=6.$ We used that

 $x+\frac{1}{x} \geqslant 2,\;x>0$, with equality if and only if $x=1$. In our case we have equality for $3ab=1$.

          To find the numbers that achieve the minimum:

$$(3a+2)(b+1)=6\Leftrightarrow 3ab+3a+2b+2=6\underset{3ab=1}{\Leftrightarrow} 3a+2b=3.$$

So we have the conditions

$$3a+2b=3\quad 3a\cdot 2b=2$$

so $3a$ and $2b$ are the roots of the quadratic equation $y^2-3y+2=0.$ It results that

$$\{3a,\;2b\}=\{1,\;2\}$$

from which we obtain the answer.

$\blacksquare$