已知正实数a、b、c满足ab+bc+ca=abc,求证:

2个回答

  • (a^4+b^4)/ab(a^3+b^3)

    =(a^4+b^4)/(a^4b+ab^4)

    因为:

    (a^4+b^4)(a+b)=(a^5+b^5)+(a^4b+ab^4)

    又(a^5+b^5)-(a^4b+ab^4)

    =(a^4-b^4)(a-b)=(a-b)^2*(a^2+b^2)(a+b)>=0

    所以(a^5+b^5)>=(a^4b+ab^4)

    (a^4+b^4)(a+b)>=2(a^4b+ab^4)=2ab(a^3+b^3)

    (a^4+b^4)/ab(a^3+b^3)>=2/(a+b)

    同样,有:

    (b^4+c^4)/bc(b^3+c^3)>=2/(b+c)

    (c^4+a^4)/ca(c^3+a^3)>=2/(c+a)

    (a^4+b^4)/ab(a^3+b^3)+(b^4+c^4)/bc(b^3+c^3)+(c^4+a^4)/ca(c^3+a^3)

    =2[1/(a+b)+1/(b+c)+1/(c+a)]

    设a>=b>=c

    则2[1/(a+b)+1/(b+c)+1/(c+a)]

    >=2[1/(2b)+1/(2c)+1/(2c)]=(1/b+2/c)

    因为ab+bc+ac=abc

    1/c+1/b+1/a=1

    1/b+2/c=1/b+1/c+1/c=1-1/a+1/c>=1

    所以式子>=1