How to solve the form $(f(x))^{1/3}>(g(x))^{1/3}$? I had a question where I had to find range of $p$ (given, it is real) in order of $(x+3p+1)^{1/3}-(x)^{1/3}=1$ to have real solutions.
I was wondering if I can take cube on both sides of inquality and solve further but then it occured to me that squaring of square roots always created problems as we weren't sure if a function was stil less than or greater than the other function.
Please help, I know this question holds equality so we can cube here but what if I had a cube root problem could I still cube it?
 A: Assuming that $h(x)=x^{\frac 1 3}$ is a real valued function defined for $x \in \mathbb R$, since $l(x)=x^{3}$ is an increasing function we have
$$f(x)^{\frac 1 3}>g(x)^{\frac 1 3}\iff \left(f(x)^{\frac 1 3}\right)^3>\left(g(x)^{\frac 1 3}\right)^3 \iff f(x)>g(x)$$
indeed also $h(x)=x^{\frac 1 3}$ is an increasing function.
Refer also to the related:

*

*cubic root of negative numbers

For the square root we have that $h(x)=x^{\frac 1 2}$ is a real valued function defined for $x \ge 0$ therefore since $l(x)=x^{2}$ is an increasing function we have
$$f(x)^{\frac 1 2}>g(x)^{\frac 1 2}\implies  \left(f(x)^{\frac 1 3}\right)^2>\left(g(x)^{\frac 1 2}\right)^2 \implies f(x)>g(x)$$
but
$$f(x)>g(x) \not \Rightarrow  f(x)^{\frac 1 2}>g(x)^{\frac 1 2}$$
because we can't exclude that $g(x)<0$.
A: Solution 1
It is not worthwhile to raise both sides of the inequality to the third power as it stands.
It is better to move one term to the other side and then raise to the third power.
$$(x+a)^{1/3}=x^{1/3}+1,\qquad a=3p+1$$
After raising to the third power we get the quadratic equation
$$3x^{2/3}+3x^{1/3}+1-a=0$$ The equation is solvable iff $$9-12(1-a)\ge 0$$ i.e. $a\ge {1\over 4}.$ Hence $p\ge -{1\over 4}.$
Solution 2
We may restrict to $a=3p+1>0,$ as the LHS of the inequality is nonpositive  for $a\le 0.$
We will apply the formula and the inequality
$$u^3-v^3=(u-v)(u^2+uv+v^2),\quad 2(u^2+uv+v^2)\ge u^2+v^2$$ Thus $$u-v\le {2(u^3-v^3)\over u^2+v^2},\qquad u>v$$
Consider the function
$$f(x)=(x+a)^{1/3}-x^{1/3}$$
Then ($u=(x+a)^{1/3},\ v=x^{1/3}$)  $$0\le f(x)\le {2a\over [(x+a)^{1/3}]^2+[x^{1/3}]^2}$$
Hence $$\lim_{|x|\to \infty} f(x)=0$$
The function $f(x)$ is decreasing on the half-axis $[0,\infty).$  This can be verified by studying the sign of the derivative on $[0,\infty).$ For $x\le 0$ the maximal value is attained at $x=-a/2.$ By the intermediate value theorem the interval $(0,4^{1/3}a^{1/3}]$ is equal to the range of $f.$ Therefore the equation has a solution if and only if $4^{1/3}a^{1/3}\ge 1,$ i.e. $p\ge -{1\over 4}.$
Remark Solution $2$ can be applied to similar question with roots of any odd order, for example $5.$ In that case the solution $1$ is convenient, if applicable at all.
