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.


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 $(x^2+y^2)^2$.

  • $\begingroup$ Nicely done. Thanks! I was going through and canceling each term out factor by factor until I got $0 \leq 6x^5\ ...$, but couldn't keep my absolute value rules in tact while doing so. $\endgroup$ – Josh Aug 12 '15 at 21:01
  • $\begingroup$ Actually, one thing, when you cancel out $(x^2+y^2)^2$, aren't you left with $\frac{|4x^3|}{x^2+y^2} + 2|x| \leq 6|x|$, and $\frac{|4x^3|}{x^2+y^2}$ is greater than $6|x|$ in some cases. $\endgroup$ – Josh Aug 12 '15 at 21:08
  • $\begingroup$ Yes but you should continue and write as I explained: $$|4x^3|=4|x|x^2\le 4|x|(x^2+y^2)$$ and then you cancel out $(x^2+y^2)$ to get $4|x|+2|x|=6|x|$. $\endgroup$ – user260717 Aug 12 '15 at 21:14
  • $\begingroup$ Yes, you have$ |4x^3| \leq 4|x|(x^2+y^2)$, but the original function has $|4x^3|(x^2 +y^2)$, so there should be an extra $(x^2+y^2)$ in there. $\endgroup$ – Josh Aug 12 '15 at 21:21
  • $\begingroup$ As I see the denominator is $(x^2+y^2)^2$ so there isn't an extra $(x^2+y^2)$. Isn't it? $\endgroup$ – user260717 Aug 12 '15 at 21:24

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.

  • $\begingroup$ Please let me know how I can improve my answer. I really just want to give you the best answer I can. $\endgroup$ – Mark Viola Aug 17 '15 at 15:08

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