已知实数x y满足√2x+1 +√2y+3=4 则x+y的最小值为?补充:根号下2x+1加根号下2y+3
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已知实数x y满足√2x+1 +√2y+3=4 则x+y的最小值为?补充:根号下2x+1加根号下2y+3
因为:(x+y)^2=x^2+y^2+2xy所以:x^2+y^2<=(x+y)^2对于本题:[根号(2x+1)]^2+[根号(2x+3)]^2<=[根号(2x+1)+根号(2y+3)]^2=4^2=16也就是:[(2x+1)+(2y+3)]<=16=>2(x+y)<=12=>x+y<=6综上,x+y的最小值为:6,当且仅当:x=-1/2;y=-3/2.解毕.补充:补充:x+y最小值的条件错了,请忽略.
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