Algebra or Calculus?

Applying Cauchy Schwarz inequality :- $(x^2+y^2+z^2)(5^2+6^2+7^2)\geq (5x+6y+7z)^2=64$ $\implies x^2+y^2+z^2\geq \dfrac{64}{110}=\dfrac{32}{55}$ $\huge\therefore ~32+55=\boxed{87}$ $\large\color{#D61F06}{\star\star\star\star\star}$ Equality occurs when: $\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}$

Geometrically we can interpret $x^2+y^2+z^2$ as the squared distance from origin of a point $(x,y,z)$ satisfying the equation of plane $\Pi: \dfrac{5x}{\sqrt{110}}+\dfrac{6y}{\sqrt{110}}+\dfrac{7z}{\sqrt{110}}=\dfrac{8}{\sqrt{110}}$ . And we know this distance is minimum for foot of perpendicular point whose distance from origin is $\left(\dfrac{8}{\sqrt{110}}\right)^2$ .

I'm going to use Lagrange multipliers .

Let $L(x,y,z) = x^2 + y^2 + z^2 + \lambda (5x + 6y + 7z - 8)$ . For finding a maximum or a minimum of $L$ is necessary $dL(x,y,z) = 0 \Rightarrow$ : $\frac{dL}{dx} (x,y,z) = 2x + 5\lambda = 0$ $\frac{dL}{dy} (x,y,z) = 2y + 6\lambda = 0$ $\frac{dL}{dz} (x,y,z) = 2z + 7\lambda = 0$ $\color{#69047E}{(*)} 5x + 6y + 7z = 8$ All this implies that $\lambda = \frac {-2x}{5} = \frac {-2y}{6} = \frac{-2z}{7}$ Substituing in the last equation we get $x = \frac{5y}{6}, \space z = \frac{7y}{6}$ and now substituing in $\color{#69047E}{(*)}$ we get $\quad y = \frac{24}{55}, \space x = \frac{4}{11}, \space z = \frac{28}{55}$ and substituing these values in $A = x^2 + y^2 + z^2$ we'll get $A = \frac{32}{55}$ . For seeing it's really a minimum, take the matrix of $d^{2} L (x,y,z)$ and see is a positive defined matrix...

Consider Two Vectors

5i+6j+7k

xi+yj+zk

Taking there dot product and setting it equal to product of their magnitudes times cosine of the angle between the two we get

32/55 = (x^2+y^2+z^2)(cosA)

Where A Is angle between two vectors

For Minimum A = 0

Note- i,j,k are unit vectors along coordinate axes

Let $\mathbf r = \begin{pmatrix} x \\ y \\ z\end{pmatrix}$ and $\mathbf a = \begin{pmatrix} 5 \\ 6 \\ 7\end{pmatrix}$ , we have $\mathbf r \cdot \mathbf a=8$ then $(\mathbf r\cdot \mathbf a) ( \mathbf r \cdot \mathbf a) = ( \mathbf r\cdot \mathbf r) (\mathbf a \cdot\mathbf a)=r^2a^2 = 64$ therefore $r^2 = 32/55$ . This is just solving the vector along the plane.

By CS-Inequality we have $( x^{2}+ y^{2}+ z^{2})( 5^{2}+6^{2}+7^{2}) \geq ( 5x+ 6y+ 7z)^{2} = 64$ . Therefore $x^{2} + y^{2}+ z^{2} \geq \frac{64}{110} = \frac{32}{55} = \frac{a}{b}$ . Thus $a+b= 87$ .