1. ## inequality

$

\begin{array}{l}
solve\;the\;inequality \\
\\
\frac{{(x + 1)^4 }}{{x(x^2 + 1)}} \succ \frac{{128}}{{15}} \\
\end{array}

$

2. First, it's important to note that for this inequality to hold, the LHS is positive. That means that either the top and bottom are both positive, or the top and bottom are both negative. I think it's easy to see that the top can never be negative...

So that means we have...

$(x + 1)^4 > 0$ and $x(x^2 + 1) > 0$

The first is true for all $x$ and the second is true only for $x > 0$. Can you see why?

So for the top and bottom to both be positive, $x > 0$.

So that means

$\frac{(x + 1)^4}{x(x^2 + 1)} > \frac{128}{15}$

$\frac{x^4 + 4x^3 + 6x^2 + 4x + 1}{x^3 + x} > \frac{128}{15}$

$15(x^4 + 4x^3 + 6x^2 + 4x + 1) > 128(x^3 + x)$ (we can cross multiply this way because we know everything is positive)

$15x^4 + 60x^3 + 90x^2 + 60x + 15 > 128x^3 + 128x$

$15x^4 - 68x^3 + 90x^2 - 68x + 15 > 0$.

Upon using the Factor Theorem

$(x - 3)(3x - 1)(5x^2 - 6x + 5) > 0$.

Can you go from here?

3. Originally Posted by Prove It
First, it's important to note that for this inequality to hold, the LHS is positive. That means that either the top and bottom are both positive, or the top and bottom are both negative. I think it's easy to see that the top can never be negative...

So that means we have...

$(x + 1)^4 > 0$ and $x(x^2 + 1) > 0$

The first is true for all $x$ and the second is true only for $x > 0$. Can you see why?

So for the top and bottom to both be positive, $x > 0$.

So that means

$\frac{(x + 1)^4}{x(x^2 + 1)} > \frac{128}{15}$

$\frac{x^4 + 4x^3 + 6x^2 + 4x + 1}{x^3 + x} > \frac{128}{15}$

$15(x^4 + 4x^3 + 6x^2 + 4x + 1) > 128(x^3 + x)$ (we can cross multiply this way because we know everything is positive)

$15x^4 + 60x^3 + 90x^2 + 60x + 15 > 128x^3 + 128x$

$15x^4 - 68x^3 + 90x^2 - 68x + 15 > 0$.

Upon using the Factor Theorem

$(x - 3)(3x - 1)(5x^2 - 6x + 5) > 0$.

Can you go from here?
4. It's not finished though. All we've found is that $x$ has to be positive, but you need to find the values it can take...
It's not finished though. All we've found is that $x$ has to be positive, but you need to find the values it can take...