Find maximize and minimize (if have) of $S=\frac{2\left(2a\right)^{2018}}{m}+\frac{2\left(2b\right)^{2018}}{n}+\frac{3c^{2018}}{p}$

Question

Let $a,b,c,m,n,p\in \mathbb{R}^+$ such that $2\sqrt[2017]{m}+2\sqrt[2017]{n}+3\sqrt[2017]{p}\le 7$ and $4a+4b+3c\le 42$. Find maximize and minimize (if have) of $$S=\frac{2\left(2a\right)^{2018}}{m}+\frac{2\left(2b\right)^{2018}}{n}+\frac{3c^{2018}}{p}$$

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I have no idea for this problem. I tried to use Holder but failed.

Answer 1

As nothing stops one among $m, n, p$ from getting arbitrarily close to $0$, $S$ is unbounded and hence there is no maximum.

For the minimum, note that we must have $4a+4b+3c=42$, as otherwise we may reduce one among $a, b, c$ to get a smaller $S$. Now using Holder’s inequality, we get $$ \left(\frac{2\left(2a\right)^{2018}}{m}+\frac{2\left(2b\right)^{2018}}{n}+\frac{3c^{2018}}{p}\right)\cdot \left(2\sqrt[2017]m+2\sqrt[2017]n+3\sqrt[2017]p \right)^{2017}\geqslant (4a+4b+3c)^{2018}$$ $$\implies S\geqslant \frac{42^{2018}}{7^{2017}}=7\cdot6^{2018}$$

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P.S. Leaving to you to confirm that equality in fact can be achieved, so this is indeed the minimum.