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Algebra Difficulty 5.3 AIME, harder Prove it Romania

a) Prove that for any real numbers aa and bb the following inequality holds:
(a2+1)(b2+1)+50≥2(2a+1)(3b+1). (a^2 + 1)(b^2 + 1) + 50 \ge 2(2a + 1)(3b + 1).

b) Find all positive integers nn and pp such that:
(n2+1)(p2+1)+45=2(2n+1)(3p+1). (n^2 + 1)(p^2 + 1) + 45 = 2(2n + 1)(3p + 1).

Solution

a) The given condition rewrites as (ab−6)2+(a−2)2+(b−3)2≥0(ab-6)^2 + (a-2)^2 + (b-3)^2 \ge 0, obviously true.

b) Similarly, we obtain (np−6)2+(n−2)2+(p−3)2=5(np-6)^2 + (n-2)^2 + (p-3)^2 = 5, hence the numbers (np−6)2(np-6)^2, (n−2)2(n-2)^2, and (p−3)2(p-3)^2 are equal to 00, 11, and 44, in some order. By inspection, we find (n,p)∈{(2,4),(2,2)}(n, p) \in \{(2, 4), (2, 2)\}.

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