+159 Let \(a, b, c, d, e, f\) be nonnegative real numbers such that \(a^2 + b^2 + c^2 + d^2 + e^2 + f^2 = 6\) and \(ab + cd + ef = 3\). What is the maximum value of \(a+b+c+d+e+f\)?
I think that you need to use Cauchy-Schwarz
+26393 Let \(a, b, c, d, e, f\) be nonnegative real numbers such that \(a^2 + b^2 + c^2 + d^2 + e^2 + f^2 = 6\) and \(ab + cd + ef = 3\).
What is the maximum value of \(a+b+c+d+e+f\) ?
The Cauchy–Schwarz inequality states that for all vectors \(\mathbf{u}\) and \(\mathbf{v}\) 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.
\(\text{Let $\vec{u} = \begin{pmatrix} a+b \\c+d\\e+f \end{pmatrix}$ } \\ \text{Let $\vec{v} = \begin{pmatrix} 1 \\1\\1 \end{pmatrix}$ } \)
\(\begin{array}{|rcll|} \hline \langle \mathbf {u} ,\mathbf {v} \rangle &=& \begin{pmatrix} a+b \\c+d\\e+f \end{pmatrix}\begin{pmatrix} 1 \\1\\1 \end{pmatrix} \\ &=& a+b+c+d+e+f \\\\ \langle \mathbf {u} ,\mathbf {u} \rangle &=& \begin{pmatrix} a+b \\c+d\\e+f \end{pmatrix}\begin{pmatrix} a+b \\c+d\\e+f \end{pmatrix} \\ &=& a^2+b^2+c^2+d^2+e^2+f^2+2(ab + cd + ef) \\ &=& 6+2(3) \\ &=&\mathbf{ 12 } \\\\ \langle \mathbf {v} ,\mathbf {v} \rangle &=& \begin{pmatrix} 1 \\1\\1 \end{pmatrix}\begin{pmatrix} 1 \\1\\1 \end{pmatrix} \\ &=& 1^2+1^2+1^2 \\ &=& \mathbf{ 3 } \\\\ \hline |\langle \mathbf {u} ,\mathbf {v} \rangle |^{2} &\leq& \langle \mathbf {u} ,\mathbf {u} \rangle \cdot \langle \mathbf {v} ,\mathbf {v} \rangle \\ (a+b+c+d+e+f)^2 &\le& 12\cdot 3 \\ (a+b+c+d+e+f)^2 &\le& 36 \\ \mathbf{ a+b+c+d+e+f } & \mathbf{\le} & \mathbf{6} \\ \hline \end{array}\)
The maximum value of \(a+b+c+d+e+f\) is \(\mathbf{6}\).
