当x,y,z>0时x+2y+3z=1u=x*x/(x+1)+4y*ysup2/(2y+1)+9z*z/(3z+1)的最小值
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当x,y,z>0时x+2y+3z=1u=x*x/(x+1)+4y*ysup2/(2y+1)+9z*z/(3z+1)的最小值当x,y,z>0时x+2y+3z=1u=x*x/(x+1)+4y*ysup2/
当x,y,z>0时x+2y+3z=1u=x*x/(x+1)+4y*ysup2/(2y+1)+9z*z/(3z+1)的最小值
当x,y,z>0时x+2y+3z=1u=x*x/(x+1)+4y*ysup2/(2y+1)+9z*z/(3z+1)的最小值
当x,y,z>0时x+2y+3z=1u=x*x/(x+1)+4y*ysup2/(2y+1)+9z*z/(3z+1)的最小值
已知x+2y+3z=1,求求u=x²/(x+1)+4y²/(2y+1)+9z²/(3z+1)最小值
根据柯西不等式,[x²/(x+1)+4y²/(2y+1)+9z²/(3z+1)]·[(x+1)+(2y+1)+(3z+1)]≥(x+2y+3z)²
∴[x²/(x+1)+4y²/(2y+1)+9z²/(3z+1)]·[(x+2y+3z)+3]≥(x+2y+3z)²
即4u≥1,从而u≥1/4
等号成立条件:[x/√(x+1)]:[√(x+1)]=[2y/√(2y+1)]:[√(2y+1)]=[3z/√(3z+1)]:[√(3z+1)]
即[x/(x+1)]=[2y/(2y+1)]=[3z/(3z+1)],[(x+1)/x]=[(2y+1)/2y]=[(3z+1)/3z]
1+1/x=1+1/2y=1+1/3z,即x=2y=3z,代入x+2y+3z=1得x=1/2,y=1/3,z=1/6
代入u后经验证u=1/4,故u最小值为1/4
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x+y+z=6 x+2y+3z=14 y+1=z
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