We know that a and B are positive real numbers, and a + B = 1, and prove that: [1 + 1 / (a ^ 2)] [1 + 1 / (b ^ 2)] is greater than or equal to 25 fast

We know that a and B are positive real numbers, and a + B = 1, and prove that: [1 + 1 / (a ^ 2)] [1 + 1 / (b ^ 2)] is greater than or equal to 25 fast

[1 + 1 / (a ^ 2)] [1 + 1 / (b ^ 2)] = [1 + A ^ 2] [1 + B ^ 2] / (AB) ^ 2 = 1 + (a ^ 2 + B ^ 2 + 1) / (AB) ^ 2 = 1 + ((a + b) ^ 2 + 1-2ab) / (AB) ^ 2 = 1 + 2 * (1-ab) / (AB) ^ 2 = 1 + 2 * (1 / (AB) ^ 2-1 / AB) = 1 + 2 [(1 / AB-1 / 2) ^ 2-1 / 4] and a + B = 1 > = 2sqrt (AB), so AB = 4, so the original formula > = 1 + 2 [(4-1 / 2) ^ 2

It is known that a, B and C belong to positive real numbers, and a + B + C = 1. It is proved that a plus one part of a multiplied by one part of B + B is greater than or equal to 25 / 4

Just a moment

Let a and B be real numbers, and 11 + a − 11 + B = 1b − a, then 1 + B1 + a equals ()
A. 1±52B. ±1+52C. ±3−52D. 3±52

Let 1 + a = x, 1 + B = y, then B-A = Y-X, the original equation can be reduced to 1 x − 1 y = 1 y − x, sorted out, y2-3xy + x2 = 0, divide both sides by x2, get (YX) 2-3 (YX) + 1 = 0, the solution is YX = 3 ± 52, that is, 1 + B1 + A is equal to 3 ± 52, so choose D