Inequality $a^2+4b^2<1$. Let $a,b$ be two strictly positive real numbers that satisfy $a^3+b^3=a-b$. How can I prove that $a^2+4b^2<1$?
So because $a^3+b^3$ is $>0$ I know that $a>b$.
I tried using $a^3+b^3=(a+b)(a^2-ab+b^2)$ but that didn‘t lead me anywhere...

Note that $$\frac {a^3+b^3 } { a-b}=1 $$
We show that $$a^2 +4b^2 <\frac {a^3+b^3 } { a-b}$$
Since $a-b$ is positive we  may multiply both sides by $a-b$ to get $$(a^2+4b^2)(a-b)< a^3+b^3$$
After simplification on both sides we get $$a^2-4ab+5b^2>0$$
That is the same as $$(a-2b)^2+b^2 >0$$ which is true.
