If $x,y>0$ and $x^2+y^3\ge x^3+y^4$, prove that $x^3+y^3 \le 2$. As in the title. If $x,y>0$ and $x^2+y^3\ge x^3+y^4$, prove that $$x^3+y^3 \le 2.$$
This seems to be a very tricky one. I tried applying various inequalities like AM-GM, unfortunately, none of techniques I'm familiar with seem to work here.
I'd greatly appreciate any hints. 

We can try to maximize $f(x,y)=x^3+y^3$ subject to
\begin{align}
&x>0\\
&y>0\\
&x^2+y^3-x^3-y^4\geq 0
\end{align}
Since $\nabla f(x,y)=(3x^2,3y^2)$, $f$ always increases away from the origin in the first quadrant. Hence, a maximum must occur on $x^2+y^3-x^3-y^4=0$.
At this point we can use Lagrange multipliers. A maximum must satisfy
\begin{align}
&x>0\\
&y>0\\
&x^2+y^3-x^3-y^4=0\\
&(3x^2,3y^2)=\lambda\cdot(2x-3x^2,3y^2-4y^3)
\end{align}
for some real $\lambda$. Then
$$(1+\lambda)3x-2\lambda=0\\3(1-\lambda)+4\lambda y=0.$$
Notice we may not have $\lambda=0$, nor $\lambda=\pm1$. Hence
$$x=\frac{2\lambda}{3(1+\lambda)}\,\,\,\,\,\,y=\frac{-3(1-\lambda)}{4\lambda}$$
Since $x$ must be positive, we find that $\lambda >0$ or $\lambda <-1$. Since $y$ must be positive, we find that $\lambda >1$ or $\lambda <0$. Hence, the only options are $\lambda>1$ or $\lambda <-1$.
Substituting $x$ and $y$ into $x^2+y^3-x^3-y^4=0$, we get a rational function in $\lambda$ whose numerator must equal $0$. It has a single root in $(-\infty,-1)\cup(1,\infty)$, at $\lambda=-3$.
With this value, we get $x=y=1$, so the maximum is attained at $(1,1)$ with precisely the value $2$.
