Double Partial Derivatives of sum of variances of dependent random variables I have the following function 
$$f(α)=Var[αX+(1−α)Y]=Var(αX)+Var[(1−α)Y]+2α(1−α)Cov(X,Y)$$

Partial derivative of this function w.r.t α leads us to following result 
$$f′(α)=2αVar(X)−2(1−α)Var(Y)+2Cov(X,Y)(1−2α)=0$$

$$ α=\frac{Var(Y)−Cov(X,Y)} {Var(X)+Var(Y)−2Cov(X,Y} $$

This α is the minimum value of the function but in order to prove that it is minimum value, one has to prove that double partial derivative of the function  f(α) is greater than Zero  i.e.  f′′(α) > 0 . 
But upon solving for f′′(α) I get

$$f′′(α)=Var(X)+Var(Y)-2Cov(X,Y)$$

How can conclude that this f′′(α) > 0 . Please help on this.
$$
0\le\def\Var{\operatorname{Var}}\Var(X-Y)=\Var(X)+\Var(Y)-2\operatorname{Cov(X,Y)}\;.
$$

Indeed this is the most you can show; you cannot show $f''(\alpha)\gt0$, since for $X=Y$ $f$ is constant.
As the variances is always a non-negative and Var(X-Y) is always greater than zero and our equaion leads to the proof that we wanted i.e. 
$$ f′′(α) &gt; 0$$
So it has been proved that α denotes the minimum value of the function.