<s>[INST] 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. 
 [/INST] Another way. Let $y=kx$.
Hence, the condition gives $x^4x^2-(k^3-1)x-1\leq0$ or
$$x\leq\frac{k^3-1+\sqrt{k^6+4k^4-2k^3+1}}{2k^4}.$$
Thus, we need to prove that
$$\left(\frac{k^3-1+\sqrt{k^6+4k^4-2k^3+1}}{2k^4}\right)^3(k^3+1)\leq2$$ or
$$3k^{12}-3k^{10}+2k^9+3k^4-2k^3+1\geq(k^9+k^7-k^6+k^4-k^3+1)\sqrt{k^6+4k^4-2k^3+1}$$ or
$$(k-1)^2(2k^{10}+4k^9-k^4-2k^3+3k^2+2k+1)\geq0.$$
Done!
Also we can make the following.
By the condition and AM-GM we obtain:
$$x^2+y^2\geq x^3+y^4+y^2-y^3\geq x^3+y^3.$$
Thus, $x\leq\frac{k^2+1}{k^3+1}$ and it's enough to prove that
$$(k^3+1)\left(\frac{k^2+1}{k^3+1}\right)^3\leq2$$ or
$$\left(\frac{k^3+1}{2}\right)^2\geq\left(\frac{k^2+1}{2}\right)^3,$$
which is just Power Means inequality.
Done!
By this way we can prove a stronger inequality: $x^5+y^5\leq2$.
Indeed, we need to prove that:
$$(k^5+1)\left(\frac{k^2+1}{k^3+1}\right)^5\leq2,$$
which is true and nice, but it's another problem.
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