Other Idea to find min of $\frac{\sqrt[3]{x-1}+\sqrt[3]{y-1}}{\sqrt[3]{x+y-2}}$

Question

The question is to find $\min\{A\}$ when $$\frac{\sqrt[3]{x-1}+\sqrt[3]{y-1}}{\sqrt[3]{x+y-2}}\leq A\\y>1,x>1$$ I tried like below $$x-1=a^3,y-1=b^3\\\frac{a+b}{\sqrt[3]{a^3+b^3}}=\frac{\frac{a}{b}+1}{\sqrt[3]{(\frac ab)^3+1}}\to \frac ab=t \to \\f(t)=\frac{t+1}{\sqrt[3]{t^3+1}}\to \\f(t)=\sqrt[3]{\frac{t^2+2t+1}{t^2-t+1}}=\\\sqrt[3]{1+\frac{3}{\underbrace{t+\frac1t}_{\geq 2}-1}}\\f_{\min}=\sqrt[3]4 $$ Now I am asking for other idea to find $\min\{A\}$, for example I can find $f'(t)=0 , t >0$ and go on. I put my second try in answers, But I am looking for the new Ideas

Answer 1

$$4(a^3+b^3)-(a+b)^3=3(a^3-a^2b-ab^2+b^3)=\cdots=3(a-b)^2(a+b)\ge0$$

Alternatively like How Do I Solve This Fascinating Equation: $\sqrt[19]{x^2+y^2-x}+\sqrt[19]{y^2-x^2+x}=2\sqrt[19]{y^2}$?, $$\left(\dfrac{a+b}2\right)^3\le\dfrac{a^3+b^3}2$$

$$\implies\dfrac{(a+b)^3}{a^3+b^3}\le4$$