設(shè)a>b>0,則a2+1ab+1a(a-b)的最小值是( ) A.1 B.2 C.3 D.4
設(shè)a>b>0,則
a2++的最小值是( ?。?br/>A. 1
B. 2
C. 3
D. 4
優(yōu)質(zhì)解答
a2++=
ab++a(a-b)+≥4
當(dāng)且僅當(dāng)
取等號(hào)
即
取等號(hào).
∴
a2++的最小值為4
故選:D
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