Cauchy like inequality $(5\alpha x+\alpha y+\beta x + 3\beta y)^2 \leq (5\alpha^2 + 2\alpha \beta +3\beta ^2)(5x^2+2xy+3y^2)$ Problem: Prove that for real $x, y, \alpha, \beta$,
$(5\alpha x+\alpha y+\beta x + 3\beta y)^2 \leq (5\alpha^2 + 2\alpha \beta +3\beta ^2)(5x^2+2xy+3y^2)$.
I am looking for an elegant (non-bashy) solution.
It closely resembles Cauchy inequality but $\alpha y +\beta y$ part is creating a problem.
I also tried to define a suitable inner product but couldn't.
Since, the inequality is homogenous, (for non-zeros) it reduces to,
$(5mn+m+n+3)^2 \leq (5m^2+2m+3)(5n^2+2n+3)$, by putting $\alpha = \beta m$ and $x=n y$. But I couldn't take it furthur from here.
So, any hints, solutions (especially along lines of inner product) would be very welcome.

By C-S $(5\alpha^2 + 2\alpha \beta +3\beta ^2)(5x^2+2xy+3y^2)=$
$=\left(\left(\sqrt5\alpha+\frac{\beta}{\sqrt5}\right)^2+\frac{14\beta^2}{5}\right)\left(\left(\sqrt5x+\frac{y}{\sqrt5}\right)^2+\frac{14y^2}{5}\right)\geq$
$=\left(\left(\sqrt5\alpha+\frac{\beta}{\sqrt5}\right)\left(\sqrt5x+\frac{y}{\sqrt5}\right)+\frac{14\beta y}{5}\right)^2=(5\alpha x+\alpha y+\beta x + 3\beta y)^2$
