Is the inequality $| \sqrt[3]{x^2} - \sqrt[3]{y^2} | \le \sqrt[3]{|x -y|^2}$ true?

Question

I'm having some trouble deciding whether this inequality is true or not... $| \sqrt[3]{x^2} - \sqrt[3]{y^2} | \le \sqrt[3]{|x -y|^2}$ for $x, y \in \mathbb{R}.$

Answer 1

$|x - y|^2 = (x - y)^2 $

Let $\sqrt[3]{x} = a , \sqrt[3]{y} = b$ I will study this $$| a^2 - b^2 | \leq \sqrt[3]{|a^3-b^3|^2} $$ L.H.S $$|a^2 - b^2|^3 = |a-b|^3 \cdot |a+b|^3= \color{red}{|a-b|^2}\cdot |a^2-b^2|\cdot |a^2 +2ab+b^2|$$ R.H.S $$|a^3 - b^3|^2 = \color{red}{|a-b|^2} \cdot |a^2 +ab + b^2|^2 $$

So our problem reduced into studying if $$|a^2-b^2|\cdot |a^2 +2ab+b^2| \leq |a^2 +ab + b^2|^2$$ If $a,b >0$. Then L.H.S $$\color{red}{a^4 +2a^3b} - 2ab^3 -b^4$$ R.H.S $$\color{red}{a^4 +2a^3b} + 3a^2b^2 +2ab^3 + b^4$$ which is absolutely bigger than L.H.S

Hope it will help you..

Answer 2

It is shown in Is $x^t$ subadditive for $t \in [0,1]$? that we have the inequality $$(a+b)^t \le a^t + b^t \tag{1}$$ for all nonnegative $a,b$ and all $t \in [0,1]$.

For your inequality, let us assume without loss of generality that $x \ge y$. Applying the above inequality (1) with $a=y$, $b=x-y$, and $t=2/3$ we get $$x^{2/3} \le y^{2/3} + (x-y)^{2/3}.$$ Subtracting $y^{2/3}$ from both sides, $$x^{2/3} - y^{2/3} \le (x-y)^{2/3}.$$ Now since $x \ge y$ we have $x-y = |x-y|$. And since the function $x \mapsto x^{2/3}$ is increasing, we have $x^{2/3} \ge y^{2/3}$, so $x^{2/3} - y^{2/3} = |x^{2/3} - y^{2/3}|$. Thus we showed $$|x^{2/3} - y^{2/3}| \le |x-y|^{2/3}$$ which is the desired inequality.

Answer 3

it boils down to showing that $|1-u^{2/3}|\leq(1-u)^{2/3}$, $1<u<1$, $u=x/y$, $0<x<y<1$.This is the same as showing that $1<u^{2/3}+(1-u)^{2/3}$. Since $0<u$, $1-u<1$ we know that $u<u^{2/3}$, $1-u<(1-u)^{2/3}$. Adding up we get the affirmative result.