Cauchy like inequality $(5\alpha x+\alpha y+\beta x + 3\beta y)^2 \leq (5\alpha^2 + 2\alpha \beta +3\beta ^2)(5x^2+2xy+3y^2)$

Question

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.

Answer 1

I assume $3 \beta^2$ is meant rather than $2\beta^2$.

This is exactly the Cauchy-Schwarz inequality for the inner product $$\langle (\alpha,\beta),(x,y) \rangle = 5\alpha x + \alpha y + \beta x + 3\beta y.$$

Answer 2

Let $A = \pmatrix{5 & 1 \\ 1 & 3}$ and for $w,v \in \mathbb R^2$ let $$(v,w)_A := v^T \cdot A \cdot w$$ be the inner product induced by $A$ (since $A$ has eigenvalues $4 \pm \sqrt{2} >0$, this indeed is an inner product). Further let $$v = \pmatrix{\alpha \\ \beta} ~ , ~ w = \pmatrix{x \\ y}.$$

Then the inequality is

$$(v,w)_A^2 \leq (v,v)_A \cdot (w,w)_A$$

which is exactly the Cauchy-Schwarz inequality.

Answer 3

By C-S $(5\alpha^2 + 2\alpha \beta +3\beta ^2)(5x^2+2xy+3y^2)=$

$=\left(\left(\sqrt5\alpha+\frac{\beta}{\sqrt5}\right)^2+\frac{14\beta^2}{5}\right)\left(\left(\sqrt5x+\frac{y}{\sqrt5}\right)^2+\frac{14y^2}{5}\right)\geq$

$=\left(\left(\sqrt5\alpha+\frac{\beta}{\sqrt5}\right)\left(\sqrt5x+\frac{y}{\sqrt5}\right)+\frac{14\beta y}{5}\right)^2=(5\alpha x+\alpha y+\beta x + 3\beta y)^2$