From weighted A.M.−G.M. inequality, we know
5+25(2x2)+2(2x5α)≥((2x2)5⋅(2x5α)2)71
⇒25x2+x5α≥27(α)72
Given that the least value of 25x2+x5αis14
i.e. 27(α)72=14⇒(α)71=2⇒α=128
If the minimum value of f(x)=25x2+x5α,x>0, is 14, then the value of α is equal to
Held on 28 Jul 2022 · Verified 6 Jul 2026.
32
64
128
256
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