已知(a+b)(a²+b²+3/2)=2,a>0,b>0 求证:a+b≤1
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已知(a+b)(a²+b²+3/2)=2,a>0,b>0 求证:a+b≤1
已知(a+b)(a²+b²+3/2)=2,a>0,b>0 求证:a+b≤1
已知(a+b)(a²+b²+3/2)=2,a>0,b>0 求证:a+b≤1
证明:a>0,b>0.
∴(a+b)(a²+b²+3/2)=2
→a+b=2/(a²+b²+3/2)
≤2/[(a+b)²/2+3/2]
→(a+b)³+3(a+b)-4≤0
→(a+b-1)[(a+b)²+(a+b)+4]≤0
→(a+b-1)[(a+b+1/2)²+15/4]≤0.
对于上式,显然,(a+b+1/2)²+15/4>0,
∴a+b-1≤0,即a+b≤1.