prove a non-strict inequality using the triangle inequality

Question

In an $\varepsilon - \ \delta $ delta proof I had to show that $$|x^3+y^3|\leq 2(x^2+y^2)^{3/2}$$ i tried some things already such as splitting it but i cannot seem to figure out how to show this. Note: i had to write from the left side untill i arrive at the right side.

Answer 1

What about squaring both sides? \begin{align*} |x^{3} + y^{3}| \leq 2(x^{2} + y^{2})^{3/2} & \Longleftrightarrow x^{6} + 2x^{3}y^{3} + y^{6} \leq 4(x^{2} + y^{2})^{3}\\\\ & \Longleftrightarrow x^{6} + 2x^{3}y^{3} + y^{6} \leq 4(x^{6} + 3x^{4}y^{2} + 3x^{2}y^{4} + y^{6})\\\\ & \Longleftrightarrow 3(x^{6} + y^{6}) + 10x^{4}y^{2} + 10x^{2}y^{4} + 2x^{2}y^{2}(x^{2} - xy + y^{2})\geq 0 \end{align*}

where the last inequality is true indeed.