Vtyjkyuk

what is the maximal value of $sqrt[3]1+(x-y)+sqrt[3]1+4(y-z)+sqrt[3]1+9(z-x)$?

I think we will use Holder inequality , try to transform the expression before using the inequality

If the expressions under the radicals can be negative then the answer is:

Does not exist.

Try $x=y=0$ and $zrightarrow+infty$.

Because for $x=y=0$ and $zrightarrow+infty$ we obtain:
$$sqrt[3]1+x-y+sqrt[3]1+4(y-z)+sqrt[3]1+9(z-x)rightarrow1+sqrt[3]1+9z-sqrt[3]4z-1=$$
$$=1+frac1+9z-(4z-1)sqrt[3](1+9z)^2+sqrt[3](1+9z)(4z-1)+sqrt[3](4z-1)^2=$$
$$=1+fracfrac2sqrt[3]z^2+sqrt[3]zsqrt[3](frac1z+9)^2+sqrt[3](frac1z+9)(4-frac1z)+sqrt[3](4-frac1z)^2rightarrow+infty.$$
If it means that $1+x-ygeq0$, $1+4(y-z)geq0$ and $1+9(z-x)geq0$, we can use the following reasoning.

Let $1+x-y=a^3$, $1+4(y-z)=8b^3$ and $1+9(y-z)=27c^3,$ where $a$, $b$ and $c$ are non-negatives.

Thus, we need to find the maximum of $a+2b+3c$, where $$a^3-1+frac8b^3-14+frac27c^3-19=0$$ or
$$36(a^3+2b^3+3c^3)=49.$$
But by Holder
$$49=36(a^3+2b^3+3c^3)=(1+2+3)^2(a^3+2b^3+3c^3)geq(a+2b+3c)^3,$$ which gives
$$a+2b+3cleqsqrt[3]49.$$
The equality occurs for $$(1,2,3)||(a^3,2b^3,2c^3),$$ which gives
$a=b=c=k$ for some value of $k$.

Thus, $$k+2k+3k=sqrt[3]49,$$ which gives
$$a=b=c=fracsqrt[3]496$$ and we got the following system:
$$1+x-y=frac49216$$ and $$1+4(y-z)=frac4927,$$ which is possible for $$(x,y,z)=left(0,frac167216,frac4172right)$$and we can say that $sqrt[3]49$ is a maximal value.