當(dāng)x,y,z>0時x+2y+3z=1u=x*x/(x+1)+4y*ysup2/(2y+1)+9z*z/(3z+1)的最小值
當(dāng)x,y,z>0時x+2y+3z=1u=x*x/(x+1)+4y*ysup2/(2y+1)+9z*z/(3z+1)的最小值
數(shù)學(xué)人氣：575 ℃時間：2020-08-28 09:59:11
優(yōu)質(zhì)解答
已知x+2y+3z=1,求求u=x²/(x+1)+4y²/(2y+1)+9z²/(3z+1)最小值根據(jù)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/(...
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