log2 1+ log2 2+ … + log2 n >= ⌊n/2」log2 (n/2) 求证明,2是底

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log2 1+ log2 2+ … + log2 n >= ⌊n/2」log2 (n/2) 求证明,2是底
log2 1+ log2 2+ … + log2 n >= ⌊n/2」log2 (n/2) 求证明,2是底

log2 1+ log2 2+ … + log2 n >= ⌊n/2」log2 (n/2) 求证明,2是底

2^(log2 1+ log2 2+ … + log2 n) =1*2*……*n = n!
2^ [(n/2)log2 (n/2)] =2^ log2 (n/2)^(n/2) = (n/2)^(n/2)
由于n!>(n/2)^(n/2)
2^(log2 1+ log2 2+ … + log2 n)>2^ (n/2)log2 (n/2)
log2 1+ log2 2+ … + log2 n >(n/2)log2 (n/2)