Winter Mathematical Competition

For $a=2$, $b=c=\frac{1}{2}$ and $n\ge 3$ the inequality is not satisfied. On the other hand, for $n=1$ it is equivalent to the AM-GM inequality. It remains to consider the case $n=2$. We shall prove that the inequality is true for $n=2$.

First solution. Set $x=bc$. Then $$abc(a^2+b^2+c^2)=ax(a^2+(b+c{)}^2-2x).$$ The function $x(p-2x)$ is increasing for $x\le \frac{p}{4}$. Since $$bc\le \frac{(b+c{)}^2}{4}\le \frac{a^2+(b+c{)}^2}{4},$$ it follows that if $b+c=b^{\prime }+c^{\prime }$ and $bc\le b^{\prime }{c}^{\prime }$, then $$abc(a^2+b^2+c^2)\le a{b}^{\prime }{c}^{\prime }(a^2+b^{\prime 2}+c^{\prime 2}).$$ Without loss of generality we can assume that $b\le 1\le c$. Set $b^{\prime }=1,c^{\prime }=b+c-1$. Since $b+c=b^{\prime }+c^{\prime }$ and $bc-b^{\prime }{c}^{\prime }=(b-1)(c-1)\le 0$, the above implies that $$abc(a^2+b^2+c^2)\le a(2-a)(a^2+1+(2-a{)}^2).$$ Set $d=(a-1{)}^2$. Then $$a(2-a)(a^2+1+(2-a{)}^2)=(1-d)(3+2d)=3-d-2d^2\le 3$$ and the given inequality (for $n=2$) follows.

Second solution. Let $n=2$ and set $ab+bc+ac=x$. Then we have $a^2+b^2+c^2=9-2x$ and $9abc\le x^2$ (this follows from the inequality $(ab+bc+ac{)}^2\ge 3abc(a+b+c)$). Hence we have to prove that $(9-2x)x^2\le 27$, which is equivalent to the obvious $(2x+3)(x-3{)}^2\ge 0$.