Minimize It!

If we consider the vectors $\mathbf{u} \; = \; \left(\begin{array}{c} 1 \\ 1/2 \\ 1/3 \\ 1/4\end{array}\right) \hspace{2cm} \mathbf{v} \; = \; \left(\begin{array}{c}4 \\ 3 \\ 2 \\ 1 \end{array}\right)$ then we see that $|\mathbf{u}| = \tfrac{1}{12}\sqrt{205}$ , $|\mathbf{v}| = \sqrt{30}$ and $\mathbf{u}\cdot\mathbf{v} = \tfrac{77}{12}$ . Thus the angle between $\mathbf{u}$ and $\mathbf{v}$ is $\theta$ , where $\cos\theta = \tfrac{77}{\sqrt{6150}}$ .

Since condition (1) is rotationally symmetric, we can rotate the vectors and consider $\hat{\mathbf{u}} \; = \; \tfrac{\sqrt{205}}{12}\left(\begin{array}{c} \cos\theta \\ \sin\theta \\ 0 \\0 \end{array} \right) \hspace{2cm} \hat{\mathbf{v}} \; = \; \sqrt{30}\left(\begin{array}{c} 1 \\ 0 \\0 \\ 0 \end{array}\right)$ and so consider the equivalent problem of minimizing $\hat{\mathbf{u}}\cdot\mathbf{x} \; = \; \tfrac{\sqrt{205}}{12}\big(x\cos\theta + y\sin\theta\big)$ subject to the conditions $|\mathbf{x}|^2 \; = \; x^2 + y^2 + z^2 + w^2 \le 1 \hspace{2cm} \hat{\mathbf{v}}\cdot\mathbf{x} \; = \; \sqrt{30}x \le 1$ and we see that this minimum is $-\tfrac{\sqrt{205}}{12}$ , achieved when $x = -\cos\theta$ , $y = -\sin\theta$ , $z=w=0$ . This makes the answer $205 + 12 = \boxed{217}$ .

I first realized inequality (2) could be completely ignored because the minimum occurs when all the variables are less than zero. So then I applied the Cauchy-Schwarz inequality with the vectors $(1,\frac{1}{2},\frac{1}{3},\frac{1}{4})$ and $(x,y,z,w)$ to get: $\Big|x+\frac{y}{2}+\frac{z}{3}+\frac{w}{4}\Big|\leq \sqrt{1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}}\sqrt{x^2+y^2+z^2+w^2}=\frac{\sqrt{205}}{12}$ This implies $x+\frac{y}{2}+\frac{z}{3}+\frac{w}{4}\geq -\frac{\sqrt{205}}{12}$ . One can verify that equality occurs if we take $(x,y,z,w)=-\frac{12}{\sqrt{205}}(1,\frac{1}{2},\frac{1}{3},\frac{1}{4})$ .