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# Cauchy-Schwarz?

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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

Aug 14, 2019

#1
+23295
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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 $$\mathbf{6}$$.

Aug 15, 2019
#2
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bravissimo!

Rom  Aug 16, 2019