been having a little trouble with this one:
The sum of two positive numbers is 8. Show that the square of one plus the cube of the other is at least 44.
Any help is appreciated
Let $\displaystyle a+b=8$ with a and b positive. Then $\displaystyle 0<a<8$ and $\displaystyle b=8-a$.
Without a loss of generality, we may choose to square $\displaystyle b$ and cube $\displaystyle a$, so:
$\displaystyle a^3+b^2$
$\displaystyle = a^3+(8-a)^2 $
$\displaystyle = a^3 + 64 - 16a + a^2 $
$\displaystyle = a^3+a^2-16a+64$
We want to find the minimum of this expression for $\displaystyle 0<a<8$.
$\displaystyle \frac{d}{da} (a^3+a^2-16a+64) = 0$
$\displaystyle \implies 3a^2+2a-16=0$
$\displaystyle \implies (3a+8)(a-2)=0$
This gives 2 critical points, but only one is in the region $\displaystyle 0<a<8$. So the minimum occurs at $\displaystyle a=2$. (You should probably show that it is in fact a minimum and not a maximum.)
Now, you need to just evaluate the expression at $\displaystyle a=2$ to determine its minimum value.