Let $ x,y,z$ be roots of the equation $ \therefore x,y,z \geq 0$ $ \therefore a = x + y + z ,b = xy + yz + zx,c = xyz$ I will prove that $ \frac {1 + a + b + c}{3 + 2a + b} - \frac {c}{b} \geq \frac {1}{3}$ It's equivalent to prove that $ ab + 2b^2 \geq 9c + 6ac$ or $ (xy + yz + xz)(x + y + z) + 2(xy + yz + zx)^2 \geq 9xyz + 6xyz(x + y + z)$ which is obviously true and hold when $ x = y = z$

this is my solution : it's easily to see that $ 1+a+b+c=(1+p)(1+q)(1+r)$ and $ 3+2a+b=(1+p)(1+q)+(1+q)(1+r)+(1+p)(1+r)$, where $ p, q$ and $ r$ are the root of the equation.therefore the problem ask to find the minimum of $ f(1+p,1+q,1+r)-f(p,q,r)$ where $ f(x,y,z)=\frac{xyz}{xy+yz+zx}.$ WLOG suppose that $ p \ge q \ge r$.so by chebychev's theorem $ 3(\frac{1}{p(1+p)}+\frac{1}{q(1+q)}+\frac{1}{r(1+r)})$ $ \ge (\frac{1}{p}+\frac{1}{q}+\frac{1}{r})(\frac{1}{1+p}+\frac{1}{1+q}+\frac{1}{1+r})$ and so : $ 3(\frac{1}{p}+\frac{1}{q}+\frac{1}{r}-\frac{1}{1+p}-\frac{1}{1+q}-\frac{1}{1+r})$ $ \ge (\frac{1}{p}+\frac{1}{q}+\frac{1}{r})(\frac{1}{1+p}+\frac{1}{1+q}+\frac{1}{1+r})$. now let $ X=\frac{1}{p}+\frac{1}{q}+\frac{1}{r}$ and $ Y=\frac{1}{1+p}+\frac{1}{1+q}+\frac{1}{1+r}$.we have $ \frac{X-Y}{XY} \ge 1/3$ so $ f(1+p,1+q,1+r)-f(p,q,r) \ge 1/3.$ equality occur when $ p=q=r$.

This problem is too easy for third question.