Maximizing $x_1+x_2+\cdots+x_{10}$, where each $x_i$ lies between $-1$ and $1$, and $x_1^3+x_2^3+\cdots+x_{10}^3=0$ 
Let $x_1, x_2, \ldots, x_{10}$ be ten quantities each lying between $-1$ and $1$, and the sum of cubes of these ten quantities is zero. Find the maximum value of $x_1+x_2+\cdots+x_{10}$.

I have tried substituting $x_1=\sin(y_1)$, $x_2=\sin(y_2)$, and so on,  and then used the identity for $\sin 3x$ to simplify the cubes leading to
$$x_1 + x_2 + \cdots x_{10} = \frac13 (\sin 3y_1 + \sin 3y_2 +\cdots + \sin 3y_{10})$$
Now is the statement
$$\frac13 (\sin 3y_1 + \sin 3y_2 +\cdots + \sin 3y_{10}) \leq \frac{10}{3}$$
right? Is the answer $10/3$? How do I maximise
$$\frac13 (\sin 3y_1 + \sin 3y_2 +\cdots + \sin 3y_{10})$$ under the constraints?

@RishiShekher What you have done is established an upper bound, to the problem that has to be attained. What the error in your argument is that you never explain whether equality can occur w.r.t the constraint, your argument is analogous to this.  We all know the following inequality is true $\left(x+\frac{1}{x}\right)\ge 2$ for all $x\in \mathbb{R}$, but the same inequality could posed as $x+\frac{1}{x}\ge 1$. The difference is the first one, equality can be attained whereas for the second it can't.
P.S: This is not an answer just a comment that points out the error, the comment box couldn't handle the word count
