First define $f(u):=(u^2-1)e^{-u^4/2},\,g(u):=(u^3+2)e^{-u^4/2}$, so the inequality we want to prove is$$\int_{\Bbb R}f(x)g(x)dx\int_{\Bbb R}f(y)g(y)dy\le\int_{\Bbb R}f^2(x)dx\int_{\Bbb R}g^2(y)dy.$$In terms of the usual inner product $\langle h_1,\,h_2\rangle:=\int_{\Bbb R}h_1(u)h_2(u)du$, this becomes $\langle f,\,g\rangle^2\le\langle f,\,f\rangle\langle g,\,g\rangle$, which is just the result we were asked to use.
First define $f(u):=(u^2-1)e^{-u^4/2},\,g(u):=(u^3+2)e^{-u^4/2}$, so the inequality we want to prove is$$\int_{\Bbb R}f(x)g(x)dx\int_{\Bbb R}f(y)g(y)dy\le\int_{\Bbb R}f^2(x)dx\int_{\Bbb R}g^2(y)dy.$$In terms of the usual inner product $\langle h_1,\,h_2\rangle:=\int_{\Bbb R}h_1(u)h_2(u)du$, this becomes $\langle f,\,g\rangle^2\le\langle f,\,f\rangle\langle g,\,g\rangle$, which is just the result we were asked to use.