problem on double integral

Let $G:[0,1] \times[0,1] \rightarrow \mathbb{R}$ be defined as$$G(t, x)=\begin{cases}t(1-x), & \text { if } t \leq x \leq 1 \\x(1-t), & \text { if } x \leq t \leq 1 \end{cases}.$$For a continuous function $f$ on $[0,1]$, define$$I[f]=\int_{0}^{1} \int_{0}^{1} G(t, x) f(t) f(x) dt\,dx$$Now i want to prove $I[f]>0$ if $f$ is not identically zero. I get by symmetry$$I[f]=2\int_{0}^{1} \int_{x}^{1} x(1-t) f(t) f(x) dt\,dx$$by reducing it i get this leaves me no where. now i dont know how i can proceed , i have thought about green functions still no steps to go ahead.