+26387 Let \(x\), \(y\), and \(z\) be real numbers such that \(x^2 + y^2 + z^2 = 1\).
Find the maximum value of \(x + y + z\).
The Cauchy–Schwarz inequality states that for all vectors \( {\displaystyle u} \) and \({\displaystyle v}\) of an inner product space it is true that
\({\displaystyle |\langle \mathbf {u} ,\mathbf {v} \rangle |^{2}\leq \langle \mathbf {u} ,\mathbf {u} \rangle \cdot \langle \mathbf {v} ,\mathbf {v} \rangle ,}\)
where \({\displaystyle \langle \cdot ,\cdot \rangle } \) is the inner product.
Source: https://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality
\(\text{Let $\vec{u} = \begin{pmatrix} x \\y\\z \end{pmatrix}$ } \\ \text{Let $\vec{v} = \begin{pmatrix} 1 \\1\\1 \end{pmatrix}$ } \)
\(\begin{array}{|rcll|} \hline \langle \mathbf {u} ,\mathbf {v} \rangle &=& \begin{pmatrix} x \\y\\z \end{pmatrix}\begin{pmatrix} 1 \\1\\1 \end{pmatrix} \\ &=& x+y+z \\\\ \langle \mathbf {u} ,\mathbf {u} \rangle &=& \begin{pmatrix} x \\y\\z \end{pmatrix}\begin{pmatrix} x \\y\\z \end{pmatrix} \\ &=& x^2+y^2+z^2 \\\\ \langle \mathbf {v} ,\mathbf {v} \rangle &=& \begin{pmatrix} 1 \\1\\1 \end{pmatrix}\begin{pmatrix} 1 \\1\\1 \end{pmatrix} \\ &=& 1^2+1^2+1^2 \\ &=& 3 \\\\ \hline |\langle \mathbf {u} ,\mathbf {v} \rangle |^{2} &\leq& \langle \mathbf {u} ,\mathbf {u} \rangle \cdot \langle \mathbf {v} ,\mathbf {v} \rangle \\ (x+y+z)^2 &\le& (x^2+y^2+z^2)\cdot 3 \\ x+y+z &\le& \sqrt{x^2+y^2+z^2}\cdot \sqrt{3} \quad | \quad x^2+y^2+z^2 = 1 \\ x+y+z &\le& 1\cdot \sqrt{3} \\ \mathbf{ x+y+z } & \mathbf{\le} & \mathbf{\sqrt{3}} \\ \hline \end{array}\)
The maximum value of \(x + y + z\) is \(\mathbf{\sqrt{3}}\)
