Prove that if $x^3 + y^3 \geq a^3$ then $x^2 + y^2 > a^2$

Question

Let $x, y, a > 0$. If $x^3 + y^3 \geq a^3$, then prove that $$x^2 + y^2 > a^2\,.$$

I have tried using the substitutions $b = \frac{x}{a}$ and $c = \frac{y}{a}$, which led to showing that if $$(b + c)^3 - 3bc(b + c) \geq 1\,,$$ then $$(b + c)^2 - 2bc > 1\,.$$ Combining the 2, led to showing that for any $b, c > 0$ we must have $1 + (b + c)(bc - 1) > 0$, at which I point I got stuck.

Answer 1

Convert to polar, letting $x=r\cos\theta$ and $y=r\sin\theta$. Then we have $$r^3(\sin^3\theta+\cos^3\theta)\ge a^3$$ Since $x,y\gt 0$, we have that $\theta\in (0,\pi/2)$. Factoring will show you that, for $\theta\in (0,\pi/2)$, the value of $\sin^3\theta+\cos^3\theta$ is always less than $1$. Thus, we have that $$r^3\gt a^3$$ and so $$r^2\gt a^2$$ Since $r^2=x^2+y^2$, we are done.

Answer 2

You can easily show by expansion and using the AM-GM Inequality that $\left(x^3+y^3\right)^2 \leq \left(x^2+y^2\right)^3$ for all $x,y\geq 0$ (the inequality is strict if $x,y>0$). Thus, if two nonnegative real numbers $x$ and $y$ satisfy $x^2+y^2\leq a^2$, then $x^3+y^3 \leq a^3$ must hold, with the equality cases $(x,y)=(0,a)$ and $(x,y)=(a,0)$.

Answer 3

Let the $p$-norm on $\mathbb R^2$ be $\|x\|_{p} = \sqrt[p]{|x|^p + |y|^p}$, for $p\ge 1$.

Then $p > q$ implies $\|x\|_{p} < \|x\|_{q}$.

Apply this to $p=3$ and $q=2$.

Answer 4

It is easy to show that \begin{eqnarray*} 3(x^2+y^2) \geq 6xy > 2xy. \end{eqnarray*} Multiply this by $x^2y^2$ and we have \begin{eqnarray*} (x^2+y^2)^3&=& x^6+3x^4y^2+3x^2y^4+y^6 \\ &>&x^6+2x^3y^3+y^6=(x^3+y^3)^2 \geq a^6. \end{eqnarray*}

Answer 5

Because everything is positive, we have that $x^3+y^3 \geq a^3$ iff $(x^3+y^3)^2 \geq (a^3)^2 = a^6$, and $x^2+y^2>a^2$ iff $(x^2+y^2)^3>(a^2)^3=a^6$. Thus, this is equivalent to showing that $$(x^3+y^3)^2 \leq (x^2+y^2)^3$$ $$x^6+y^6+2x^3y^3 \leq x^6+y^6+x^4y^2+x^2y^4$$ $$2x^3y^3 \leq x^4y^2+x^2y^4$$ $$2xy \leq x^2+y^2$$ $$0 \leq x^2+y^2-2xy$$ $$0 \leq(x-y)^2$$

Since any real number squared is at least 0, this holds.