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...
 A: 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.
A: Here is a less clever (but always working) method:
Lets maximize $F(x,y)=x^2+4y^2$ under the constrains $x,y\ge 0$ and $x^3+y^3-x+y=0$ using Lagrange multipliers. Let $f(x,y,\lambda)=F(x,y)+\lambda(x^3+y^3-x+y),$ then $$\dfrac{\partial f}{\partial x}=2x+3\lambda x^2-\lambda,\,\,\,\,\,\,\dfrac{\partial f}{\partial y}=8x+3\lambda y^2+\lambda ,\,\,\,\,\,\,\dfrac{\partial f}{\partial\lambda}=x^3+y^3-x+y$$ and we get a highly non-linear system of equations. With the help of wolframalpha we can have very accurate approximate solutions (Exact solutions are bit complicated). Now for $$x\approx 0.9725956862081514773 ,\,\,\,\,\,\, y\approx 0.0524320766715842$$ we get $$F(x,y)\approx 0.95693885948708459008$$ which is a very good approximation of the maximum of $F.$ 
