Maximum $x_1$ -coordinate

Let $P$ be the set of all 20-vectors $(x_1, x_2, x_3, ...,x_{20})$ such that the $x_i's$ are real numbers and $x_1+ x_2+ x_3+ ...+x_{20}=0$ and the 20-vector $\vec{a}=\frac{1}{\sqrt{20}}(1, 1, 1,...,1).$ Clearly, $\|\vec{a}\|=1,$ and $\vec{u}\in P$ if and only if $\vec{u}\cdot\vec{a}=0.$ Let $\vec{e_1}$ be the 20-vector $(1, 0, 0, ..., 0),$ then the vector $\vec{v}=\vec{e_1}-(\vec{e_1}\cdot \vec{a})\vec{a} \in P,$ and then $\vec{e_1}=(\vec{e_1}\cdot \vec{a})\vec{a} +\vec{v} \quad \quad \quad (*)$

Now, let's considered an arbitrary 20-vector $\vec{u}=(x_1, x_2, ..., x_{20})$ in $P$ satisfying the condition that $\|\vec{u}\|=x_1^2+ x_2^2+ x_3^2+ ...+x_{20}^2=1.$ Then using (*), we have that $x_1=\vec{e_1}\cdot\vec{u}= ( (\vec{e_1}\cdot \vec{a})\vec{a} +\vec{v} )\cdot \vec{u}=\vec{v}\cdot\vec{u}.$ Therefore, the maximum value of $x_1$ is attained when $\vec{u}=\frac{\vec{v}}{\|\vec{v}\|}$ and the maximum value will be $\vec{v}\cdot\frac{\vec{v}}{\|\vec{v}\|}=\|\vec{v}\|.$

By definition of $\vec{v},$ we have that $\vec{v}=(\frac{19}{20}, -\frac{1}{20}, -\frac{1}{20}, ..., -\frac{1}{20}).$ Then $m=\|\vec{v}\|=\sqrt{\frac{19}{20}} \approx \boxed{0.97}$ .

We want to maximize the function $F(x_1,x_2,...,x_{20}) = x_1$ with the two given constraints. Applying the method of Lagrange Multipliers we would have:

$(1,0,...,0) = \mu (1,1,...,1) + \lambda (2x_1,2x_2,...,2x_{20})$ . This vector equality implies that for $i > 1$ , $2\lambda x_i + \mu = 0 \implies x_i = - \frac{\mu}{2 \lambda}$ . But this implies that $x_2 = x_3 = ... = x_{20}$ . For simplicity, let $y = x_2$ . Then we have $x_1 + 19y = 0 \\ x_1^2 + 19y^2 = 1$ Which implies $x_1 ^2 = 1 - 19y^2 \\ x_1 ^2 = 19^2 y ^2$ So $1 - 19y^2 = 19^2 y^2 \implies y^2 (19^2 + 19 ) = 1 \implies y^2 = \frac{1}{380}$ . With this value we can plug in $x_1 ^2 = \frac{19^2}{380} = 0.95$ .