Advanced Algebra Manipulation/Inequality Proof: $\frac{4x^3(x^2+y^2)-2x(x^4+y^4)}{(x^2+y^2)^2} \leq 6|x|$ I need to show that
$$\frac{4x^3(x^2+y^2)-2x(x^4+y^4)}{(x^2+y^2)^2} \leq 6|x|$$
by starting with the left side of the inequality and working from there.
Hints from the textbook said to work from these inequality "tricks,"
$$2|ab| \leq a^2 + b^2$$
$$|a| + |b| \leq \sqrt{2} \sqrt{a^2+b^2}$$
And the triangle inequality,
$$|a+b| \leq |a| + |b|$$
Expanding the numerator gives us 
$$\frac{2x^5 + 4x^3y^2 - 2xy^4}{(x^2 + y^2)^2}$$
For starters, I use the triangle inequality theorem to make the numerator look like this:
$$|2x^5 + 4x^3y^2 - 2xy^4| \leq |2x^5|+|4x^3y^2|+|2xy^4|$$
Also, in regard to the denominator, since it is even, it is always positive and thus equal to its absolute value. Then, I multiply both sides by the denominator, $(x^2+y^2)^2$. This gives us
$$|2x^5|+|4x^3y^2|+|2xy^4| \leq 6|x||x^4 + 2x^2y^2+y^4|$$
$$|2x^5|+|4x^3y^2|+|2xy^4| \leq 6|x^5+2x^3y^2+xy^4|$$
My question is, since $|a+b| \leq |a|+|b|$, I feel as though I cannot violate that rule and am stuck. What trick(s) can I use from here? If there were no absolute value symbols, it would be as simple as simplifying the inequality since the orders match up. However, it would not be true for all (x,y) as the right side of the inequality is odd.
 A: We have
$$|4x^3(x^2+y^2)-2x(x^4+y^4)|\le4|x|x^2(x^2+y^2)+2|x|(x^2+y^2)^2\\\le6|x|(x^2+y^2)^2$$
and then the result is immediate by canceling the factor $(x^2+y^2)^2$.
A: Interestingly, this inequality is not tight.  In fact, we can show that 
$$\left|\frac{4x^3(x^2+y^2)-2x(x^4+y^4))}{(x^2+y^2)^2}\right|\le 2|x|$$
To that end, we write
$$\begin{align}
\frac{4x^3(x^2+y^2)-2x(x^4+y^4))}{(x^2+y^2)^2}&=\frac{2x}{(x^2+y^2)^2}\left(2x^2(x^2+y^2)-(x^4+y^4)\right)\\\\
&=\frac{2x}{(x^2+y^2)^2}\left(2(x^2+y^2-y^2)(x^2+y^2)\\
\,\,\,\,\,\,\,\,\,\,-((x^2+y^2)^2-2x^2y^2)\right)\\\\
&=2x\left(1-\frac{y^4}{(x^2+y^2)^2}\right) \tag 1
\end{align}$$

It is easy to show that the term in parentheses on the right-hand side of $(1)$ satisfies the inequality 
$$0\le \left(1-\frac{y^4}{(x^2+y^2)^2}\right)\le 1$$
We simply note that $x^2+y^2\ge y^2$ so that $0\le\dfrac{y^4}{(x^2+y^2)^2}\le 1$ and therefore $1\ge 1-\dfrac{y^4}{(x^2+y^2)^2}\ge 0$.

Thus, we have for $x>0$ 
$$0<2x\left(1-\frac{y^4}{(x^2+y^2)^2}\right)\le 2x \tag 2$$
while for $x<0$, we have
$$2x\le 2x\left(1-\frac{y^4}{(x^2+y^2)^2}\right)<0 \tag 3$$
Putting $(1)$, $(2)$, and $(3)$ together yields
$$\bbox[5px,border:2px solid #C0A000]{\left|\frac{4x^3(x^2+y^2)-2x(x^4+y^4))}{(x^2+y^2)^2}\right|\le 2|x|}$$
which provided a tighter inequality than that one that was asked to be shown.
