Pregătire Matematică Olimpiade Juniori // Junior Mathematics Olympiad Preparation // آمادگی برای المپیاد ریاضی نوجوانان

                                             Andrei ECKSTEIN has a website "Mathematics Preparation Junior Olympics"

(Pregătire Matematică Olimpiade Juniori). From the  4th Cyprus Preselection, 09/05/2026 let's train with the Problem :

     "For all positive real numbers  $a,\;b,\;c$  with  $a+b+c=3$ , prove that

$\frac{bc}{a^3}+\frac{ca}{b^3}+\frac{ab}{c^3}\geqslant 3$"

ANSWER CiP

The  $"="$  sign appears iff  $a=b=c=1$

                    Solution CiP

          The problem is simple and only requires applying the  AM-GM Inequality  for three variables :

$\frac{x+y+z}{3}\geqslant \sqrt[3]{xyz}\;\;\;,\;\;\;x,\;y,\;z\geqslant 0\;\;;\;\;"="\;iff\;x=y=z \tag{1}$

     First, we have : $\frac{a+b+c}{3}\geqslant \sqrt[3]{abc}\;\;\Rightarrow$

$\frac{1}{\sqrt[3]{abc}}\geqslant \frac{3}{a+b+c} \tag{2}$

     If we write  (1) for  $x=\frac{1}{a^4}\;,\;y=\frac{1}{b^4}\;,\;z=\frac{1}{c^4}$  we obtain

$\frac{1}{3}\cdot \left ( \frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4}\right )\geqslant \sqrt[3]{\frac{1}{a^4}\frac{1}{b^4}\frac{1}{c^4}}=\frac{1}{abc\sqrt[3]{abc}}$ , so

$\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4}\geqslant \frac{3}{abc\sqrt[3]{abc}} \tag{3}$

Now the problem is solved simply :

$\frac{bc}{a^3}+\frac{ca}{b^3}+\frac{ab}{c^3}=abc\cdot \left (\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4} \right )\underset{(3)}{\geqslant}abc\cdot \frac{3}{abc\sqrt[3]{abc}}=3\cdot \frac{1}{\sqrt[3]{abc}}\underset{(2)}{\geqslant}3\cdot \frac{3}{a+b+c}\overset{a+b+c=3}{=}\frac{9}{3}=3$

     The  $"="$  sign appears iff  $a=b=c\;for\;(2)$  and  $\frac{1}{a^4}=\frac{1}{b^4}=\frac{1}{c^4}\;for\;(3)$ , so  $a=b=c=\frac{a+b+c}{3}=1$

$\blacksquare$