Cauchy Schwarz: Proving this inequality

Question

For positive reals a,b,x and y, prove that $(5a^2 + 2ab + 3b^2)(5x^2 + 2xy + 3y^2) \ge (5ax + ay + bx + 3by)^2$

My attempt: By Cauchy we can show that the LHS $\ge(5ax + 2(abxy)^{1/2} + 3by)^2$

I can then show that $ay + bx \ge 2(abxy)^{1/2}$ using AM-GM, but this doesn't help.

How can I proceed? Thank you.

Answer 1

By C-S we obtain: $$(5a^2+2ab+3b^2)(5x^2+2xy+3y^2)=\left(\frac{1}{5}(5a+b)^2+\frac{14}{5}b^2\right)\left(\frac{1}{5}(5x+y)^2+\frac{14}{5}y^2\right)\geq$$ $$\geq\left(\frac{1}{5}(5a+b)(5x+y)+\frac{14}{5}by\right)^2=(5ax+ay+bx+3by)^2.$$

Answer 2

Expanding all, collecting linke terms,and factorization gives: $$(ay-bx)^2\geq 0$$ which is true.

Answer 3

By Gauss composition, see Dirichlet's description on page 49 of the second edition of David A. Cox, Primes of the Form $x^2 + n y^2$

your product equals $$ \left( \; 5ax+ay+bx+3by \; \right)^{ \; 2} + 14 (ay-bx)^2 $$