Prove equation with norm in inner product space

Question

$x,y$ - elements of inner product space : $ \|x+2y\|^2=5$ and $ \|2x+y\|^2=4$. Prove that $9\|x+y\|^2+\|x-y\|^2=18 $ Could I ask for any tips how should I start this exercise? I cannot use the fact that norm becomes from scalar product because it's not a Hilbert space, I've thought about using parallelogram law but I don't see how.

Answer 1

Let $u=x+y$ and $v=x-y$ then $x+2y=(3u-v)/2$ and $2x+y=(3u+v)/2$. Hence $$5=\|(3u-v)/2\|^2=\frac{9}{4}\|u\|^2-\frac{3}{2}<u,v>+\frac{1}{4}\|v\|^2,$$ and $$4=\|(3u-v)/2\|^2=\frac{9}{4}\|u\|^2+\frac{3}{2}<u,v>+\frac{1}{4}\|v\|^2.$$ Finally add the two equations: $$9=4+5=\frac{9}{2}\|u\|^2+\frac{1}{2}\|v\|^2,$$ that is $$18=9\|x+y\|^2+\|x-y\|^2.$$