Relationship between variance and covariance

Question

I know that $$\text{var}(x-y) = \text{var}(x) + \text{var}(y) - 2\text{cov}(x,y)$$ and $$\text{cov}(x,y) = \frac{1}{2}(\text{var}(x) + \text{var}(y) - \text{var}(x-y)).$$

Is it possible to say that

$$\text{cov}(x,y) \leq \text{var}(x) + \text{var}(y)$$

If I plug this inequality in above equations, I get that $-\text{var}(x-y) \leq \text{var}(x) + \text{var}(y)$, which is true since variance is positive.

Answer 1

$$\text{cov}(X,Y) = \frac{1}{2}(\text{var}(X) + \text{var}(Y) - \text{var}(X-Y))\implies \text{cov}(X,Y) \leq \frac{1}{2}(\text{var}(X) + \text{var(Y)})\leq \text{var(X)}+\text{var}(Y),$$ with equality holding iff $\text{var}(X-Y)=0$ and $\text{var}(X)+\text{var}(Y)=0$, that is, iff $$X=\text{constant}, Y=\text{constant}$$