Solutions
1. A straightforward, if inelegant, approach is to multiply out and expand everything. All terms cancel except four and we are left with 2abcd ≤ a^{2}d^{2} + b^{2}c^{2}, which is obviously true since (ad - bc)^{2} ≥ 0.
2. Differentiating gives f '(x) = (2 + x^{2})^{1/2}(3 + x^{3})^{1/3} + terms with factor (1 + x). Hence f '(-1) =3^{1/2}2^{1/3}.
3. Let the ray AB' meet CD at X and the ray BA' meet CD at Y. If AB' and A'B intersect, then X = Y. Let L be the line through A' parallel to CD. Then L is perpendicular to AA'. Hence CD is perpendicular to AA'. Similarly, let L' be the line through B' parallel to CD. Then L' is perpendicular to BB', and hence CD is perpendicular to BB'. So CD is perpendicular to two non-parallel lines in the plane ABX. Hence it is perpendicular to all lines in the plane ABX and, in particular, to AB.
Suppose conversely that AB is perpendicular to CD. Consider the plane ABY. CD is perpendicular to AB and to AA', so CD is perpendicular to the plane. Similarly CD is perpendicular to the plane ABX. But it can only be perpendicular to a single plane through AB. Hence X = Y and so AA' and BB' belong to the same plane and therefore meet.
4. Put a = sin^{2}x, b = cos^{2}x. Then a and b are non-negative with sum 1, so we may put a = 1/2 + h, b = 1/2 - h. Then a^{3} + b^{3} = 1/4 + 3h^{2} ≥ 1/4 with equality iff h = 0. Hence x is a solution of the equation given iff sin^{2}x = cos^{2}x = 1/2 or x is an odd multiple of π/4.
Source: Nguyễn Thị Lan Phương, http://321math.blogspot.com
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