$$V(aX+bY)=a^2V(X)+2ab \space\text{Cov}(X,Y)+b^2V(Y).$$ Now, althoug $X$ and $Y$ are not constant (so $V(X)>0,V(Y)>0$), it is possible for $aX+bY$ to be constant, in wich case $V(aX+bY)$ would be zero.Therefor $$V(aX+bY)\ge0,$$thus $$a^2V(X)+2ab \space\text{Cov}(X,Y)+b^2V(Y)\ge 0.$$
I was hoping someone could explain this to me. It is to do with proof that the range of correlation is between $1$ and $-1.$
I am not sure what it means to claim that the random variables $X$ and $Y$ are not constant? Does this simply mean not all values are the expected value of $Y$ or $X$ because the variance is given by
$$V(N) = E(n - E(n))^2$$
Additionally, why is it possible for $aX + bY$ to be constant?
Would very much appreciate anyone's insight.