11-th Czech-Slovak-Polish Match, 2011

Adding three AM-GM inequalities
$$4a^3b + b^3c + 2c^3a \geq 7a^2bc,$$
$$4b^3c + c^3a + 2a^3b \geq 7b^2ca,$$
$$4c^3a + a^3b + 2b^3c \geq 7c^2ab$$
we get
$$a^3b + b^3c + c^3a \geq a^2bc + b^2ca + c^2ab \quad (1)$$
The assumption $a^2 < bc$ implies $-a^3b > -b^2ca$ and this together with (1) gives
$$b^3c + c^3a > a^2bc + c^2ab$$
i.e.
$$b^3 + ac^2 > ab(a + c)$$
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