Indija TS 2008

The Cauchy-Schwarz inequality gives $$(a+b+c{)}^2\le 3(a^2+b^2+c^2).$$ Thus $(a+b+c)<6$ and hence $\frac{(a+b+c{)}^3}{9}<4(a+b+c)$. Now the AM-GM inequality gives $$(a+b+c{)}^3\ge 27abc.$$ Thus $$3abc\le \frac{(a+b+c{)}^3}{9}<4(a+b+c).$$

Alternately, we have $a+b+c<6$ so that $a^2+b^2+c^2<2(a+b+c)<12$. Thus $$(a^2+b^2+c^2)(a+b+c)<12(a+b+c).$$ Using the AM-GM inequality, we have $$a^2+b^2+c^2\ge 3(abc{)}^{2/3},a+b+c\ge 3(abc{)}^{1/3}.$$ Thus $$9abc\le (a^2+b^2+c^2)(a+b+c)<12(a+b+c),$$ giving the required inequality. Thus