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Consider the following quantities $x_{1},x_{2},x_{3},\cdots \cdots ,x_{10}$

and $-1\leq x_{1},x_{2},x_{3},\cdots \cdots ,x_{10}\leq 1.$ and

$x^3_{1}+x^3_{2}+\cdots+x^{3}_{10}=0.$

Then maximum of $x_{1}+x_{2}+x_{3}+\cdots\cdots+x_{10}$

Try : Let $x_{i} = \sin(a_{i})$ for $i = 1$ to $10$ and $\displaystyle a_{i}\in \bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg]$

So we have given

$\sin^3(a_{1})+\sin^3(a_{2})+\cdots \cdots+\sin^3(a_{10})=0$

and we have to find maximum value of

$\sin(a_{1})+\sin(a_{2})+\cdots+\sin(a_{10}).$

I did not understand how to calculate it.

Could some help me, thanks

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DXT
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  • Is the idea of writing $x_j$ as the sine of some $a_j$ yours or a hint from the book / teacher...? I ask to know if I should follow that approach or look for another one. – ajotatxe Dec 14 '18 at 13:57
  • Leaving x3 to x10 frixed, consider only x1, x2. Then the problem is transformed to maximaize x_1+x_2 under the condition that x1^3 + x2^3 is given. – Euduardo Dec 14 '18 at 13:58
  • It is mine because $x_{i}\in [-1,1].$ – DXT Dec 14 '18 at 13:58
  • @ Euduardo I did not understand why we fix $x_{3}$ to $x_{10}$ , please explain me – DXT Dec 14 '18 at 14:04
  • @DurgeshTiwari To make the problem smaller and easier. In fact, try fixing $x_4 = \cdots = x_{10} = 0$ and find the maximum of $x_1+x_2+x_3$ given $x_1^3 + x_2^3 + x_3^3 = 0$ (you need three variables to have anything interesting happen). – Arthur Dec 14 '18 at 14:30

1 Answers1

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Let us transform the variables using $f(t)=\sqrt[3]t$, to equivalently maximise $ \sum f(x_k)$ subject to $\sum x_k = 0$ and $x_k \in [-1, 1]$, where index $k\in \mathbb N, 1\leqslant k\leqslant 10$. WLOG using symmetry we may assume the $x_k$ are in non-descending order.

Noticing $f(x)$ is convex for $x< 0$, if the first $n$ among the $x_k$ are the negatives, Karamata's Inequality gives: $$\sum_{k\leqslant n} f(x_k)\leqslant (n-1)f(-1) + f\left(\sum_{k\leqslant n}x_k +n-1\right)$$ Hence we may replace the first $n$ (i.e. all negative) terms with $n-1$ copies of $-1$ and some number, say $p^3 = \sum_{k\leqslant n}x_k +n-1$.

Now the remaining $10-n$ terms are non-negative, a domain where $f$ is concave, so again by Karamata or Jensen, we may replace them all by their arithmetic mean (say $q^3$) for a sum which is at least as high.

Now consider the numbers $(p^3, q^3)$, which obviously obey $-1 \leqslant p$ and $q>0$.
$\bullet \quad$ If $p^3+q^3 \geqslant 0$, note $f(p^3)+f(q^3)\leqslant 2f(\frac12p^3+\frac12q^3) \iff p+q\leqslant 2\sqrt[3]{\frac{p^3+q^3}2} \iff 3(p-q)^2(p+q)\geqslant 0$, so we may replace both by a non-negative number.
$\bullet \quad$ OTOH, if $p^3+q^3< 0$, then $f(p^3)+f(q^3) \leqslant f(-1)+f(1+p^3+q^3) \iff p+q \leqslant -1+\sqrt[3]{1+p^3+q^3}\\ \iff -3(1+p)(1+q)(p+q)\geqslant 0$, so we may replace $(p^3, q^3)$ with $(-1, 1+p^3+q^3)$ for at lease as high a sum. Note that $1+p^3+q^3\geqslant 0$

In both cases, we are left with copies of $-1$ and non-negative numbers, which after another application of Jensen, leave us with $x_k \in \{-1, a\}$ for some $a>0$.

Thus we have for some natural $m \in [10]$ and some real $a \in (0, 1)$, $$\sum_k f(x_k) \leqslant mf(-1) + (10-m)f(a) = -m+(10-m)\sqrt[3]a$$ subject to $(10-m)a=m \implies a = \frac m {10-m}$. So we are left to find the natural $m$ that maximises $-m+(10-m)\sqrt[3]{\frac{m}{10-m}} = \sqrt[3]{m(10-m)^2}-m$.

Using calculus (extending $m$ to reals and checking both the integers around the unimodal maximum of $m=\frac{10}9$) or an easy enumeration, it is found $m=1$ is optimal choice, for a maximum of $9^{2/3}-1=3\sqrt[3]3-1$.

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