Question

设x>0,y>0,且x2+y2/2=1，求x*跟下1+y2的最大值

Answer 1

x2+y2/2=1,求x√1+y2最大值

x²+y²/2=1
2x²+y²=2
2x²+y²+1=3
即2x²与(1+y²)的和为定值
[x*√(1+y²)]²
=x²*(1+y²)
=(1/2)*2x²*(1+y²)
≤(1/2)*[2x²+(1+y²)]²/4
=(1/8)*9
=9/8
∴竖颤x*√(1+y²)≤√(9/8)=(3√2)/4
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基本不等式：ab≤(a+b)²/逗历4,当a=b时取余指败等号

Answer 2

直接算可能更方便些

x2+(y2) ／2=1
2x2+y2=2
2x2+y2+1=3
2x2+y2+1≥2√[2x2（y2+1）]= 2√2·x√﹙y2+1﹚
3≥2√2·x√﹙y2+1﹚搜烂
3√2／4≥液镇x√﹙y2+1﹚闹漏粗

Answer 3

x2+y2/2=1,求x√1+y2最大值
x²+y²/2=1
2x²+y²=2
2x²+y²+1=3
即2x²与(1+y²)的和为定值
[x*√(1+y²)]²
=x²*(1+y²)
=(1/2)*2x²*(1+y²竖颤)
≤(1/2)*[2x²+(1+y²)]²/4
=(1/8)*9
=9/8
∴x*√(1+y²)≤√(9/8)=(3√2)/4
---------------------------------
基本不余指败等式:ab≤(a+b)²/逗历4,当a=b时取等号