Poly or Ineq?

$\begin{aligned} \frac {x^4+3x^2+1}{x^2} + \frac {x^4-4x^2+1}{x^3+x} & = 4 \\ x^2 + 3 + \frac 1{x^2} + \frac {(x^2+1)^2-6x^2}{x(x^2+1)} & = 4 \\ x^2 +3 + \frac 1{x^2} + \frac {x^2+1}x - \frac {6x}{x^2+1} & = 4 \\ x^2 -1 + \frac 1{x^2} + x + \frac 1x - \frac {6}{x+\frac 1x} & = 0 \\ \left(x + \frac 1x\right)^2 + x + \frac 1x - 3 - \frac {6}{x+\frac 1x} & = 0 & \small \color{#3D99F6} \text{Let }y = x + \frac 1x \\ y^2 + y - 3 - \frac 6y & = 0 \\ y^3 + y^2 - 3y - 6 & = 0 \\ (y-2)(y^2+3y+3) & = 0 \\ \implies y & = 2 & \small \color{#3D99F6} y^2+3y+3 \text{ has no real root.} \\ x + \frac 1x & = 2 \\ \implies x & = \boxed{1} \end{aligned}$

I've seen solutions from @Rahil Sehgal and @Chew-Seong Cheong, and both of them solved this problem the Poly way, so I think that I should post an Ineq solution for the sake of completeness (if this solution is wrong, the Poly way is the only way). Please check out for any mistakes!

So it's easily seen that $x = 0$ must not be the solution, else LHS is undefined.

We can see that $\frac { { x }^{ 4 }+3{ x }^{ 2 }+1 }{ x^{ 2 } } =1+\frac { ({ x }^{ 2 }+1)^{ 2 } }{ { x }^{ 2 } } =(\frac { { x }^{ 2 }+1 }{ x } )^{ 2 }+1\ge 4+1=5$ .

Also, $\frac { { x }^{ 4 }-4{ x }^{ 2 }+1 }{ x^{ 3 }+x } =1+\frac { ({ x }^{ 2 }+1)^{ 2 } }{ { x(x }^{ 2 }+1) } -\frac { 6{ x }^{ 2 } }{ x({ x }^{ 2 }+1) } =\frac { { x }^{ 2 }+1 }{ x } -\frac { 6x }{ { x }^{ 2 }+1 } \ge 2-3=-1$ .

$=> \frac { { x }^{ 4 }+3{ x }^{ 2 }+1 }{ x^{ 2 } } + \frac { { x }^{ 4 }-4{ x }^{ 2 }+1 }{ x^{ 3 }+x } \ge 5-1=4$ .

Equality holds iff (if and only if) $x = 1$ .

And if anyone sees this, I use $x^{2} + 1 \ge 2x$ and $x \neq 0$ to solve the problem. Any flaws? (Found one)

Edit: Seems like this can only be true if $x > 0$ . So we will have to prove that $x > 0$ for this solution to be complete. Any help?

$\dfrac { x^{ 4 }+3{ x }^{ 2 }+1 }{ { x }^{ 2 } } +\dfrac { { x }^{ 4 }-4{ x }^{ 2 }+1 }{ { x }^{ 3 }+x } = 4$ .

$\Rightarrow \dfrac { (x^{ 4 }+3{ x }^{ 2 }+1)(x^2+1) +( { x }^{ 4 }-4{ x }^{ 2 }+1)(x) }{ x^4 + x^2 } = 4$

$\Rightarrow \color{#3D99F6}{(x^6 + 3x^4 + x^2 + x^4 + 3x^2 +1)} + \color{#D61F06}{(x^5-4x^3+x)} + \color{#69047E}{(-4x^4 -4x^2 )} =0$

$\Rightarrow x^6 + x^5 -4x^3 +x +1 =0$

$\Rightarrow \color{#20A900}{(x-1)^2} \color{#EC7300}{(x^4+3x^3 +5x^2 +3x+1)} \color{#333333}{=0}$

Therefore, the answer is $\color{#20A900}{x= \boxed{1}}$