Show that $A$ is non-negative definite If  $A=\begin{bmatrix}1&\dfrac{1}{2} &\dfrac{1}{3}\\\dfrac{1}{2} &\dfrac{1}{3}&\dfrac{1}{4}\\\dfrac{1}{3} &\dfrac{1}{4}&\dfrac{1}{5}\end{bmatrix}$

Show that $A$ is non-negative definite i.e. $v^tAv\ge 0$.Also prove it for a $n\times n$ matrix where the entries are given following pattern in $A$.

My try:
I did take $v=(v_1,v_2,v_3)$ and found that $v^TAv=v_1^2+v_1v_2+\dfrac{v_2^2}{3}+\dfrac{v_3^2}{5}+\dfrac{v_2v_3}{2}+\dfrac{2v_1v_3}{3}$.
I can't conclude that it is non-negative .Neither can I prove it for an $n\times n$ matrix.
Please help me .
 A: We have that $${\bf v}^tA{\bf v} = v_1^2 + v_1v_2+{v_2^2\over3} + {v_3^2\over5} + {v_2v_3}+{2\over3}v_1v_3$$
Let $v_1 = r\cos(\theta)$, $v_2=r\sin(\theta)$, and $v_3 = z$. Then we have that $$\begin{align}{\bf v}^tA{\bf v} &= v_1^2 + v_1v_2+{v_2^2\over3} + {v_3^2\over5} + {v_2v_3}+{2\over3}v_1v_3 \\ &= r^2\underbrace{\left({2\over3}+{1\over2}\sin(2\theta)+{1\over3}\cos^2(\theta)-{1\over3}\sin^2(\theta)\right)}_{\ge{6\over100} \; \text{when} \; \theta \approx 5.20379}+ rz\underbrace{\left({1\over2}\sin(\theta)+{2\over3}\cos(\theta)\right)}_{\ge-{1\over10} \text{ when } \theta\approx5.20379} + {z^2\over5} \\&\ge{6\over100}r^2-{1\over10}rz+{z^2\over5}\\&= {3\over50}\left(r-{5\over6}z\right)^2+{19\over120}z^2 \\ &\ge 0,\end{align}$$ as desired.
A: First, clearly the matrix is symmetric, but also all its principal minors are all three positive:
$$\begin{align*}&|A_{11}|=1>0\\{}\\&|A_{22}|=\begin{vmatrix}1&\frac12\\\frac12&\frac13\end{vmatrix}=\frac13-\frac14=\frac1{12}>0\\{}\\
&|A|=\begin{vmatrix}1&\frac12&\frac13\\
\frac12&\frac13&\frac14\\\frac13&\frac14&\frac15\end{vmatrix}=60^3\begin{vmatrix}60&30&20\\30&20&15\\20&15&12\end{vmatrix}\stackrel{R_1-3R_3,R_2-\frac32R_3}=\begin{vmatrix}0&-15&-16\\0&-\frac52&-3\\20&15&12\end{vmatrix}=\begin{vmatrix}15&16\\5&6\end{vmatrix}>0\end{align*}$$
and thus the matrix is even positive definite, not only no-negative. Observe that in the last line we only care about the determinant's sign, not its actual value: in the second step we must multiply by $\;60^3\;$, and in the last one by $\;20\;$ (we developed by the first column), but these are positive constants so we don't care.
