已知正数a,b,c,x,y,z满足a+x=b+y=c+z=k,求证ay+bz+cx<k^2答:由条件可知a+x=k,①b+y=k,②c+z=k,③将三个式子两边分别平方,并相加得3k^2=(a^2+x^2)+(b^2+y^2)+(c^2+z^2)+2(ax+by+cz),利用不等式a^2+b^2≥2ab,3k^2≥2ax+2by+2cz+2(ax+by+cz)=4(ax+by+cz),所以k^2≥4(ax+by+cz)/3>...
已知正数a,b,c,d,e,f满足答:acdef/b=9...② abdef/c=16...③ abcef/d=1/4...④ abcdf/e=1/9...⑤ abcde/f=1/16...⑥ ①×②×③×④×⑤×⑥得:a^5b^5c^5d^5e^5f^5/abcdef=1 (abcdef)^4=1 因为a,b,c,d,e,f都是正数 得abcdef=1 === 如果将abcdef=1带入①可得a^2=1/4,a=1/2 同...