Question

Find which is the greater $3a{b}^2$ or $a^3+2b^3$.

Answer 1

let $X=a^3+2b^3;Y=3a{b}^2$

let us consider X-Y;

$=a^3+2b^3-3a{b}^2$

$=a^3-a{b}^2-2a{b}^2+2b^3$

$=a(a^2-b^2)-2b^2(a-b)$ ;$(a^2-b^2)=(a+b)(a-b)$

$=(a-b)(a^2+ab-b^2)$

let $a>b$;Say $X-Y>0$

$=>(a-b)(a^2+ab-b^2)>0$

$=>(a-b)\left[\right(a+\frac{b}{2}{)}^2-\frac{9b^2}{4}]>0$

Here since we assumed $a>b;a-b>0$

$=>(a+\frac{b}{2}{)}^2-\frac{9b^2}{4}>0$

$=>(a+\frac{b}{2}{)}^2>\frac{9b^2}{4}$

$=>a>b$ which is the correct condition as we assumed.

Hence our assumption $X-Y>0$ is correct and hence $a^3+2b^3>3a{b}^2$