如何用综合法证明不等式?用综合法证明,若a>0,b>0,则（a³+b³）/2≥[(a+b)/2]³

Question

如何用综合法证明不等式?
用综合法证明,若a>0,b>0,则（a³+b³）/2≥[(a+b)/2]³

Answer 1

a>0,b>0
(a-b)^2(a+b)≥0
a^3+b^3-a^2b-ab^2≥0
a^3+b^3≥a^2b+ab^2
3a^3+3b^3≥3a^2b+3ab^2
4a^3+4b^3≥a^3+b^3+3a^2b+3ab^2
4(a^3+b^3)≥(a+b)^3
(a^3+b^3)/2≥(a+b)^3/8
(a^3+b^3)/2≥[(a+b)/2]^3