If $x,y>0$ and $x^2+y^3\ge x^3+y^4$, prove that $x^3+y^3 \le 2$.

Question

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.

Answer 1

By AM-GM $2x^3+1\geq3x^2$ and $3y^4+1\geq4y^3$.

Thus, $$2x^3+3y^4+2\geq3x^2+4y^3$$ or $$x^3+y^3\leq2+3(x^3+y^4-x^2-y^3)\leq2$$

Answer 2

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 $2$.

Answer 3

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.