已知a>0,b>0,且a+b=1.求证;根号下a+1⼀2加上根号下b+1⼀2

已知a>0,b>0,且a+b=1.求证;根号下a+1/2加上根号下b+1/2<=2
2024-11-25 17:37:27
推荐回答(2个)
回答1:

(根号(a+1/2)+根号(b+1/2))²<=2(a+1/2+b+1/2)=2×(1+1)=4
开方得
根号(a+1/2)+根号(b+1/2)<=2

基本不等式:(x+y)²<=2(x²+y²)
证明很简单,略

回答2:

由:(x-y)²>=0得x²+y²>=2xy,两边同加上x²+y²可得基本不等式:(x+y)²<=2(x²+y²) 两边开根号→|x+y|<=√2(x²+y²)
用上面的式子可得(√a+1/2)+(√b+1/2)<=√2(a+1/2)^2+2(b+1/2)^2=√2*(a+b+1)=2