The Problem "S:L23.322" and probably its neighbors

             

    It is page 9 of the "SUPLIMENTUL cu EXERCIȚII al GMB N0 12/2023".

          In translation
               "S:L23.322.    Show that if  $\;a,\;b,\;c\in (0,\;\infty)$  and  $a+b+c=1$, then

$$\frac{a+1}{\sqrt{a+bc}}\geqslant 2."$$

           Solution CiP 

           We have the known inequality 

$\frac{x+y}{\sqrt{xy}}\geqslant 2 \tag{1}$

where $x,\;y$ are positive real numbers. The sign $"="$ occurs in (1) if and only if $x=y.$

(The deduction of (1) is immediate from the equivalents

$$(\sqrt{x}-\sqrt{y})^2 \geqslant 0\;\Leftrightarrow\;x-2\sqrt{x}\cdot \sqrt{y}+y \geqslant 0\;\Leftrightarrow\;x+y\geqslant 2\sqrt{xy}\;\Leftrightarrow\;(1)\;)$$

          Back to the problem, it results from the given conditions that $\;a,b,c<1$ (because, for example $a=1-b-c<1-0-0$ and $-b,-c<0$).

Then the numbers  $x=1-b$  and  $y=1-c$  are positive. But we have
$$a+1=1-b-c+1=(1-b)+(1-c)=x+y$$

and

$$a+bc=1-b-c+bc=(1-b)(1-c)=xy$$

and then the inequality (1) expresses exactly what the statement requires.

$\blacksquare$

          Remark CiP  The statement does not ask when the $"="$ sign occurs. We will answer, as a bonus.

The sign $"="$ occurs if and only if $(a,b,c)=(1-2t,t,t)$ with  $t\in (0, \frac{1}{2}).$ 

Even the values $t=0$ (when $a=1,\;b=c=0$) and $t=\frac{1}{2}$ (when $a=0,\;b=c=\frac{1}{2}$) are admissible for the statement.

<end Rem>

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          As Mr. Konstantine ZELATOR used to us, a problem never appears alone. I am referring to his  posts related to problems in the Crux Mathematicorum magazine (although he also has concerns about UFO-logy). As I announced in the title, we will also try to solve the other problems on the photographed page.

 

              " S:L23.321.  Let $ABCD$ be the convex quadrilateral whose diagonals 

          intersect at $O$. If $\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{AO}=\overrightarrow{BC}+\overrightarrow{DC}+\overrightarrow{OC}$, show that $ABCD$

          is a parallelogram."

                  Solution CiP

          It seems that no figure would be needed.

I did one, with a trembling hand.

          The relationship given between the six vectors is written, replacing some of them with the sum of two others, in the form

$(\overrightarrow{AO}+\overrightarrow{OB})+(\overrightarrow{AO}+\overrightarrow{OD})+\overrightarrow{AO}=(\overrightarrow{BO}+\overrightarrow{OC})+(\overrightarrow{DO}+\overrightarrow{OC})+\overrightarrow{OC}\;\Leftrightarrow$

$\Leftrightarrow\;3\overrightarrow{AO}+\overrightarrow{OB}+\overrightarrow{OD}=\overrightarrow{BO}+3\overrightarrow{OC}+\overrightarrow{DO}\;\Leftrightarrow$

$\Leftrightarrow \;3(\overrightarrow{AO}-\overrightarrow{OC})=2\overrightarrow{BO}+2\overrightarrow{DO}\;\Leftrightarrow\;$

$\Leftrightarrow\;3(\overrightarrow{AO}+\overrightarrow{CO})=2(\overrightarrow{BO}+\overrightarrow{DO}).\;\tag{1}$

     On the left side of the relation (1) we have a sum of two vectors  collinear with $\overrightarrow{AC}$, so the result will also be a vector collinear with $\overrightarrow{AC}$. On the right side of the relation (1) we have a sum of two  vectors collinear with $\overrightarrow{BD}$, so the result will also be a vector collinear with $\overrightarrow{BD}$. Thus, equality (1) is possible only if

$$\overrightarrow{AO}+\overrightarrow{CO}=\overrightarrow{0}=\overrightarrow{BO}+\overrightarrow{DO}\;\Leftrightarrow$$

$\Leftrightarrow\;\overrightarrow{AO}=-\overrightarrow{CO}=\overrightarrow{OC}$ and $\overrightarrow{BO}=-\overrightarrow{DO}=\overrightarrow{OD}.$ But this means that point $O$ is both the midpoint of segment $[AC]$ and the midpoint of $[BD]$, thus $ABCD$ is a parallelogram.

$\blacksquare$