I am interested in proving the following sub version of Cauchy Schawrz equality.
1) LA version :
If $x$ and $y$ are two real vectors and the following holds $$<x,y> = ||x||.||y||$$ then $x$ and $y$ are proportional to each other
2) Expectation version :
If $X$ and $Y$ are two non-negative random variables and the following holds $$ E[XY] = \sqrt{E[X^2]E[Y^2]}$$ then $X$ and $Y$ are proportional to each other
Proof:
1) LA version :
We know that $x$ and $y$ are proportioanl to each other iff the unit vector along them are equal i.e.
$$\frac{x}{||x||}=\frac{y}{||y||}$$
So, if we can prove that $x||y||-y||x||=\theta_v$ then we are done i.e. if we can prove that inner product of $x||y||-y||x||$ with itself is zero then we are done. If we find the above inner product and put the given equally then the inner product becomes zero.
Now, my question is is there any similar "intuitive" proof exists for the expectation version or we have to put a special $\lambda$ (so this is a trick !) in the expression of $E[X-\lambda Y]^2$. Hope I have made my point.
I think you can pretty much copy your argument from the linear algebra situation into the expectation case. Define $$ Z=X\sqrt{E(Y^2)}-Y\sqrt{E(X^2)}, $$ and notice that $$ Z^2=X^2E(Y^2)+Y^2E(X^2)-2XY\sqrt{E(X^2)}\sqrt{E(Y^2)}, $$ so taking expectations of both sides we get $$ E(Z^2)=2E(X^2)E(Y^2)-2E(XY)\sqrt{E(X^2)}\sqrt{E(Y^2)}. $$ Now if you assume $E(XY)=\sqrt{E(X^2)}\sqrt{E(Y^2)}$, then this simplifies to $E(Z^2)=0$. But the only way a non-negative random variable like $Z^2$ can have expectation $0$ is to be $0$ almost surely. That is, with probability $1$, $X\sqrt{E(Y^2)}=Y\sqrt{E(X^2)}$, which makes the random variables $X$ and $Y$ proportional with probability $1$.
(You can't get rid of "with probability $1$ and say they're proportional everywhere, since your assumption $X\sqrt{E(Y^2)}=Y\sqrt{E(X^2)}$ isn't changed when you modify $X$ and $Y$ on sets of measure zero.)