x+2y+3z+4w=1,求x2+y2+z2+w2+(x+y+z+w)2最小值

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x+2y+3z+4w=1,求x2+y2+z2+w2+(x+y+z+w)2最小值

x+2y+3z+4w=1,求x2+y2+z2+w2+(x+y+z+w)2最小值
x+2y+3z+4w=1,求x2+y2+z2+w2+(x+y+z+w)2最小值

x+2y+3z+4w=1,求x2+y2+z2+w2+(x+y+z+w)2最小值
x+2y+3z+4w=1,依权方和不等式得
x²+y²+z²+w²+(x+y+z+w)²
=x²/1+(3y)²/9+(5z)²/25+(7w)²/49+(x+y+z+w)²/1
≥[x+3y+5z+7w+(x+y+z+w)]²/(1+9+25+49+1)
=[2(x+2y+3z+4w)]²/85
=4/85.
故所求最小值为:4/85.