İkisi Bir Arada: Yağ ve su gibi karışmazlar // Two in One : they don't mix, like oil and water

 Strange juxtaposition of two issues in SGMB 3/2025

In translation:

     "Let  $a,\;b,\;c$  be positive real numbers. Show that

$\frac{\sqrt{a}}{b+c}=\frac{\sqrt{b}}{a+c}=\frac{\sqrt{c}}{a+b} \tag{E}$

                                  if and only if  $a=b=c.$"

                                        Solution CiP

            Let's assume that  $a\neq b$. Then, from the first two of the equalities (E), we get

$0<\frac{\sqrt{a}}{b+c}=\frac{\sqrt{b}}{a+c}=\frac{\sqrt{a}-\sqrt{b}}{(b+c)-(a+c)}=\frac{\sqrt{a}-\sqrt{b}}{b-a}=-\frac{\sqrt{a}-\sqrt{b}}{a-b}=$

$=-\frac{\sqrt{a}-\sqrt{b}}{(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}=-\frac{1}{\sqrt{a}+\sqrt{b}}<0$

but this is ABSURD. 

          So $a=b$, and therefore from the equalities (E) there remains

$$\frac{\sqrt{a}}{a+c}=\frac{\sqrt{c}}{2\cdot a}$$

Assuming now that  $c\neq a$, the above results in

$0<\frac{\sqrt{a}}{a+c}=\frac{\sqrt{a}-\sqrt{c}}{(a+c)-2a}=-\frac{\sqrt{a}-\sqrt{c}}{a-c}=-\frac{1}{\sqrt{a}+\sqrt{c}}<0$

this is again ABSURD. So  $c=a=b$.

QED $\blacksquare$