Question

Prove ab(a+b) <= a^3 + b^3 if a,b >= 0

* <= less than or equal to
* >= greater than or equal to

thanks

Answer 1

a^3 +b^3=(a+b)(a^2-ab+b^2)
then:

Answer 2

\[a^2+b^2\geq2\sqrt{a^2b^2}=2ab\]
\[a^2-ab+b^2\geq ab\]
\[a^2-ab+b^2\geq ab/\cdot(a+b)\]
\[(a^2-ab+b^2)\cdot(a+b)\geq ab\cdot(a+b)\]
\[a^3+b^3\geq ab(a+b)\]

Answer 3

ab(a+b)<=(a+b)(a^2-ab+b^2)=0
ab<=a^2-ab+b^2
(a-b)^2>=0
true for any a& b