How do i prove this inequality? Let $a,b,c,d,e,f$ be positive reals.

Then how do i prove the below inequality?


  If $ac-b^2≧0$ and $df-e^2≧0$, then $(a+d)(c+f)-(b+e)^2 ≧0$.


I hope someone answers this specifically.. Thank you in advance. :)
Another way: 

consider quadratic polynomials $f(x)=ax^2+b/2 x+c$ and $g(x)=dx^2+e/2x+f$. Then your inequalities are equivalent to the negativity of discriminants of $f$ and $g$. And this is equivalent  to $f(x) \geq 0$ and $g(x) \geq 0$ for all $x$. But then $(f+g)(x) \geq 0$ for all x. So the discriminant of $f+g$ is negative. This is precisely $(a+d)(c+f)-(b+e)^2 \geq 0$
Given that $ac - b^2 \geq 0$ and $df - e^2 \geq 0$
.We can rewrite it as $ac\geq b^2$ and $df\geq e^2\quad\quad (1)$

We have $(a+d)(c+f) = ac+df+dc+af$. 
The first two terms are respectively $\geq b^2$ and $e^2$ using $(1)$.

So, we have $ac + df \geq b^2 + e^2\quad \quad (2)$.

Since $a,b,c,d,e,f$ are positive reals, we can apply the AM-GM inequality as below:
$$\frac{af+dc}{2}\geq \sqrt{acdf}$$ 
In other words $$af + cd \geq 2\sqrt{acdf}\quad\quad (3)$$

Also, $\sqrt{acdf}\geq be$ using $(1)$ again .......$(4)$.

$(3)$ and $(4)$ gives $$af+dc\geq 2\sqrt{acdf}\geq 2be\quad\quad (5)$$

$(5)$ and $(2)$ gives your required result.