Help writing proof for $\sqrt{a^2 + b^2} \neq \sqrt[3]{a^3 + b^3}$

Solution 1:

Hint: Take the sixth power of both sides.

Added: (for completeness, after the OP had used the hint to solve the problem) Since $\sqrt{a^2+b^2}$ is by definition non-negative, we have $$\sqrt{a^2+b^2}=\sqrt[3]{a^3+b^3} \qquad\text{iff}\qquad \left(\sqrt{a^2+b^2}\right)^6=\left(\sqrt[3]{a^3+b^3}\right)^6.$$ Equivalently, we want to find the solutions of $$(a^2+b^2)^3=(a^3+b^3)^2.$$ in non-zero reals. Expand. We get $$a^6 +3a^4b^2+3a^2b^4+b^6=a^6+2a^3b^3+b^6,$$ which is equivalent to $$3a^4b^2+3a^2b^4=2a^3b^3.$$ Since we are looking for solutions where neither $a$ nor $b$ is equal to $0$, we are looking for non-zero real solutions of $$3a^2+3b^2=2ab.$$ This equation has no non-zero real solutions. For by completing the square we get $$a^2+b^2-\frac{2ab}{3}=\left(a-\frac{b}{3}\right)^2+\frac{8b^2}{9}.$$ In order for the right hand side to be $0$, both terms must be $0$. In particular, $b$ must be $0$, and therefore so must $a$.

Comment: You may find the following related idea interesting. Let $p>1$, and let $t$ be positive. We will prove that $$1+t> (1+t^p)^{1/p}. \qquad\qquad(\ast)$$ Let $f(t)=1+t -(1+t^p)^{1/p}$. Note that $f(0)=0$. So it is enough to show that for positive $t$, $f(t)$ is an increasing function. To do this, we use the derivative: $$f'(t)=1 -\frac{t^{p-1}}{(1+t^p)^{(p-1)/p}}.$$ For positive $t$, $f'(t)$ is positive, since $(1+t^p)^{(p-1)/p}>(t^p)^{(p-1)/p}=t^{p-1}$. This completes the proof of $(\ast)$.

To apply $(\ast)$ to our problem, observe first that a solution of our equation with positive $a$ yields a solution of $(1+(b/a)^2)^{1/2}=(1+(b/a)^3)^{1/3}$. It is easy to check that $b/a$ cannot be negative. Now let $t=(b/a)^2$. We obtain the equation $1+t=(1+t^{3/2})^{2/3}$, which, by $(\ast)$, cannot hold for positive $t$.

We reduced the problem to one variable in order to use familiar tools. But there are important generalizations to several variables.