Least value of $b$ for which inequality is true. 
find the least natural number $b$ for which $x+bx^{-2}>2,\forall \space x\in (0,\infty)$

What I got confused with
In my book they have done this: rearranged the function as $$f(x)=x^3-2x^2+b$$ an $f(x)>0$
They have told to find least value of $b$, $f(x)$ should be minimum. This is what i didn't understand. Why should $f(x)$ be minimum $\implies$ $b$ is minimum.
 A: You can get the function $f(x)$ by multiplying both sides by $x^2$, and moving everything to the lhs, which gives $f(x)>0$. It is "hardest" for this inequality to hold at a minimum, in the sense that if $f(x)>0$ holds at a minimum, it must hold for any other $x'$, since at any $x'$, $f(x')\geq f(x)>0$. Minimizing this function gives the FOC
$$3x^2-4x=0$$
which is solved at either $x=4/3$ or $x=0$. Checking the second order conditions gives us 
$$6x-4,$$
so $x=4/3$ is a local minimum. Moreover, $f(x)$ is increasing to the right of $4/3$, so it is a global minimum. 
Finally, the inequality holds as long as $f(4/3)>0$, so we are looking for the smallest natural number such that
$$(4/3)^3-2(4/3)^2+b>0$$
So, $b$ needs to solve
$$b>32/27,$$
which means that 2 is the least natural number where this holds everywhere. 
A: $$
g(x) = x + b/x^2 > 2 \quad (x > 0) \quad (*) \iff \\
x^3 + b > 2 x^2 \quad (x > 0) \iff \\
f(x) = x^3 - 2 x^2 + b > 0 \quad (x > 0) \quad (**)
$$
A $b$ that will keep $(*)$ true, will also keep $(**)$ true and vice versa.
Thus we are are talking the same set
$$
B = \{ b \mid g(x) > 2, x > 0 \} = \{ b \mid f(x) > 0, x > 0 \}
$$
here. 
To find such a $b$ we are looking for a minimum of $f$ on $(0, \infty)$:
$$
0 = f'(x) = 3x^2 - 4 x
$$
which vanishes there for $x = 4/3$, the other candidate $x=0$ was outside our interval of interest. Because $f''(x) = 6x - 4$ and $f''(4/3) = 24/3 - 4 = 4 > 0$ it is a minimum. 
The value of $f$ there is
$$
f(4/3) 
= (4/3)^3 - 2(4/3)^2 + b 
= 64/27 - 32/9 + b 
= 64/27 - 96/27 + b 
= -32/27 + b
$$
If we adjust it to touch the $x$-axis we get $b = 32/27 = 1.19$ so we choose $b = 2$.
Because the minimum of $f$ is positve, the other values of $f$ on $(0,\infty)$ (which are larger) will be too. So $f$ at the minimum location was the most difficult customer.
A: Given $$x+\frac{b}{x^2}>2\;\forall\; x>0$$
Here $x>0\;,b>0$
Using $\bf{A.M\geq G.M\;,}$ We get
$$\frac{x}{2}+\frac{x}{2}+\frac{b}{x^2}\geq 3\left(\frac{x}{2}\cdot \frac{x}{2}\cdot \frac{b}{x^2}\right)^{\frac{1}{3}} = 3\left(\frac{b}{4}\right)^{\frac{1}{3}}>2$$
So we get $$b>\frac{32}{27}$$
So we get $\min$ natural no. $b=2$
