Let $A=\left(\begin{matrix}\frac{1}{1+1}&\frac{1}{1+2}&\cdots&\frac{1}{1+n}&\\\frac{1}{2+1}&\frac{1}{2+2}&\cdots&\frac{1}{2+n}\\\vdots&\vdots&\ddots&\vdots\\\frac{1}{n+1}&\frac{1}{n+2}&\cdots&\frac{1}{n+n}\end{matrix}\right)$
Can we prove that all eigenvalues of $A$ are positive?
This will follow from a general result for inner product space. Let $v_1,v_2,..v_n$ are n vectors in an inner product space $V$. Then the matrix $(\langle v_i,v_j\rangle)_{1\leq i,j\leq n }$ is semi positive definite.
Now you take $V$ to be $C[0,1]$ with the inner product $\langle f,g\rangle=\int_0^1 fg \,dx $ . Then in the space if you take $v_i=x^{i-1/2},1\leq i\leq n$ you will get the result that the above matrix is semi positive definite. Hence it has positive eigen values.
To prove the general result let $c_1,c_2,..,c_n$ are any real numbers. Then $$\sum_{i,j=1}^{n}c_ic_j\langle v_i,v_j\rangle=\langle \sum_{i=0}^{n}c_iv_i,\sum_{i=0}^{n}c_iv_i\rangle\geq 0 $$
