已知A=(1+P)(1+q),B=(1+二分之p+q)的平方,试比较A,B的大小.
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已知A=(1+P)(1+q),B=(1+二分之p+q)的平方,试比较A,B的大小.
已知A=(1+P)(1+q),B=(1+二分之p+q)的平方,试比较A,B的大小.
已知A=(1+P)(1+q),B=(1+二分之p+q)的平方,试比较A,B的大小.
A-B=1+p+q+pq-1-(p+q)-(p+q)²/4
=pq-(p²+2pq+q²)/4
=-(-4pq+p²+2pq+q²)/4
=-(p²-2pq+q²)/4
=-(p-q)²/4≤0
A≤B
A=(1+p)(1+q)=1+pq+p+q
B=[1+(p+q)/2]²=1+(p+q)+(p+q)²/4
所以B-A=(p+q)²/4-pq=(p²+2pq+q²)/4-pq=(p²-2pq+q²)/4=(p-q)²/4≥0
所以B≥A