Question

不等式数学题,设a.b为实数,a^2+2b^2=6,则a+b的最小值是?

Answer 1

换元,可设a=(√6)cosx,b=(√3)sinx.(x∈R).则a+b=(√6)cosx+(√3)sinx=3sin(x+t).(t为锐角,且sint=(√6)/3.cost=(√3)/3.)即a+b=3sin(x+t).由-1≤sin(x+t)≤1.可知,-3≤a+b≤3.故(a+b)min=-3.选C.