Macedonian Junior Mathematical Olympiad

By the trivial inequality $(x-y{)}^2+(y-z{)}^2+(z-x{)}^2\ge 0$ we have that for every $x$, $y$, $z\in \mathbb{R}$ it holds $$x^2+y^2+z^2\ge xy+yz+zx.$$ By applying this inequality, we get $$\begin{array}{rl}{a}^4+b^4+c^4& \ge a^2{b}^2+b^2{c}^2+c^2{a}^2=(ab{)}^2+(bc{)}^2+(ca{)}^2\\ & \ge (ab)(bc)+(bc)(ca)+(ca)(ab)=abc(a+b+c)=a+b+c\end{array}$$ which finishes the proof. Equality holds if and only if $a=b=c=1$.