A weighted geometric inequality

Question

$x,y,z$ are positive reals. for a triangle of side lengths $ a, b, c$ , and area $A$, the following inequality holds: $$(xa^2+yb^2+zc^2)^2\ge 16(xy+yz+zx)A^2$$ This is like a weighted Weitzenböck's inequality. I hope to get some tips to prove this question

Answer 1

Let $a^2=q+r$, $b^2=p+r$ and $c^2=p+q$.

Thus, by C-S twice we obtain: $$xa^2+yb^2+zc^2=x(q+r)+y(p+r)+z(p+q)=$$ $$=(x+y+z)(p+q+r)-xp-yq-zr=$$ $$=\sqrt{(x^2+y^2+z^2+2(xy+xz+yz))(p^2+q^2+r^2+2(pq+pr+qr))}-xp-yq-zr\geq$$ $$\geq\sqrt{(x^2+y^2+z^2)(p^2+q^2+r^2)}+2\sqrt{(xy+xz+yz)(pq+pr+qr)}-xp-yq-zr\geq$$ $$\geq2\sqrt{(xy+xz+yz)(pq+pr+qr)}=$$ $$=\sqrt{(xy+xz+yz)\sum_{cyc}(a^2+b^2-c^2)(a^2+c^2-b^2)}=$$ $$=\sqrt{(xy+xz+yz)\sum_{cyc}(2a^2b^2-a^4)}=4\sqrt{xy+xz+yz}A.$$ Done!