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.