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

$|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..
