Given the function f (x) = xlnx + (4-x) ln (4-x), if a > 0, b > 0, it is proved that alna + blnb ≥ (a + b) LNA + B2

Given the function f (x) = xlnx + (4-x) ln (4-x), if a > 0, b > 0, it is proved that alna + blnb ≥ (a + b) LNA + B2

∵ f (x) = xlnx + (4-x) ln (4-x), ∵ f ′ (x) = lnx-ln (4-x) = lnx4 − X. when x = 2, the function f (x) has the minimum value. A > 0, b > 0, let a + B = 4, then alna + blnb = alna + (4-A) ln (4-A) ≥ 2 · a + b2ln (a + B2) = (a + b) LNA + B2. ∵ alna + bln

Elden Ring Premiere

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