已知函数f(x)=x^2/2x+1(x>0)当x1>0,x2>0且f(x1)*f(x2)=1求证:x1*x2>=3+2倍
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已知函数f(x)=x^2/2x+1(x>0)当x1>0,x2>0且f(x1)*f(x2)=1求证:x1*x2>=3+2倍根号2
x1^2*x2^2=4x1x2+2(x1+x2)+1
≥4x1x2+4√(x1x2)+1
令t=√x1x2
t^4-4t^2-4t-1≥0
(t^4-2t^2+1)-2(t^2+2t+1)≥0
(t^2-1)^2-2(t+1)^2≥0
((t+1)(t-1))^2-2(t+1)^2≥0
(t+1)^2((t-1)^2-2)≥0
显然(t+1)^2≥0
所以(t-1)^2-2≥0
t-1≥√2
t≥√2 +1
√x1x2≥√2 +1
x1x2≥3+2√2
≥4x1x2+4√(x1x2)+1
令t=√x1x2
t^4-4t^2-4t-1≥0
(t^4-2t^2+1)-2(t^2+2t+1)≥0
(t^2-1)^2-2(t+1)^2≥0
((t+1)(t-1))^2-2(t+1)^2≥0
(t+1)^2((t-1)^2-2)≥0
显然(t+1)^2≥0
所以(t-1)^2-2≥0
t-1≥√2
t≥√2 +1
√x1x2≥√2 +1
x1x2≥3+2√2
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