Baltic Way Shortlisted Problems, 2015

Setting $x = \frac{a+b}{2}$ and $y = \frac{a-b}{2}$, we have $a = x + y$ and $b = x - y$. The given equations transform into
$$c^2 = x^2 + \frac{y^2}{3} \quad (1)$$
$$d^3 = x(x^2 + y^2), \quad (2)$$
and the inequality to be proved into $2x + d \le 3c$.

By (1), we have $c \ge x > 0$. Moreover, a simple calculation shows
$$(3c - 2x)^2 = 9c^2 - 12cx + 4x^2 = 3c^2 + 6(c-x)^2 - 2x^2 \ge 3c^2 - 2x^2 = x^2 + y^2,$$
using (1) in the last step.
Next, we observe that (2) implies $d \ge x$ and hence also $x^2 + y^2 \ge d^2$. In combination with $3c - 2x \ge c > 0$, this leads to
$$3c - 2x \ge \sqrt{x^2 + y^2} \ge d,$$
whereby the problem is solved.

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