$X>0$, $Y>0$ and $X^2+Y ^3\geq X^3+Y^4$. Prove that $X^3+Y ^3\leq 2$

Question

$X>0$, $Y>0$ and $X^2+Y ^3\geq X^3+Y^4$. Prove that $X^3+Y ^3\leq 2$

First I tried:

$0<X\leq1$ and $0<Y\leq1$, but this in not the only case for $X^2+Y ^3\geq X^3+Y^4$ and get nowhere.

Can somobody give me an idea?

Answer 1

The case $$a=b$$ is easy to solve. Let now $$a>b$$ then we have $$a^2+b^3\geq a^3+b^4>b^3+b^4$$ and we get

$$a>b^2$$ Now we have also two cases: $$a>b\geq b^2$$ then we get $$b\le 1$$, then it must be $$1-a\geq 0$$ and so we have $$a^3+b^3\le 2$$ Now let $$a>b^2\geq b$$ from here we get $$0\geq 1-b$$ and it must be $$a>1$$ and it follows $$0>1-a$$ and this can not be, since we have $$a^2(1-a)+b^3(1-b)\geq 0$$ The case $$a<b$$ is for you!

Answer 2

Adding $y^3$ to both sides of the inequality we get:

$$x^2 + 2y^3 -y^4\ge x^3 + y^3$$ $$x^2 + y^2 - y^2(1-y)^2 \ge x^3 + y^3$$

Thus we get that $x^2 + y^2 \ge x^3 + y^3$

However by the Root Mean Inequality we have:

$$\sqrt[3]{\frac{x^3 + y^3}{2}} \ge \sqrt{\frac{x^2+y^2}{2}}$$ $$2(x^3 + y^3)^2 \ge (x^2 + y^2)^3 \ge (x^3 + y^3)^3 $$ $$2 \ge x^3 + y^3$$

Hence the proof.