Problem 1064

Expanding the left-hand side of the inequality term yields $a b + 2 a + 2 b + 4 \geq c d$. Given the condition $a^{2} + b^{2} + c^{2} + d^{2} = 4$, the inequality can be equivalently written as

$$a b + 2 a + 2 b + a^{2} + b^{2} + c^{2} + d^{2} \geq c d$$

which, after multiplying by 2, becomes:

$$2 a^{2} + 2 b^{2} + 2 c^{2} + 2 d^{2} + 2 a b + 4 a + 4 b - 2 c d \geq 0$$

Since the left-hand side expression can be represented as

$$\begin{aligned} a^{2} + b^{2} + 2 a b + 4 a + 4 b + a^{2} + b^{2} & + c^{2} + d^{2} + (c - d)^{2} = \\ & = a^{2} + b^{2} + 2 a b + 4 a + 4 b + 4 + (c - d)^{2} = (a + b + 2)^{2} + (c - d)^{2} \end{aligned}$$

the inequality is proven. Equality holds precisely when

$$a + b = -2 \quad \text{and} \quad c = d \quad \text{with} \quad a^{2} + b^{2} + c^{2} + d^{2} = 4$$

which, in particular, is true for $a = b = -1$ and $c = d = 1$.

(Walther Janous)