Can someone help me prove this integrals inequality
$$\int_0^1\sqrt{f^4(x)+\bigg(\int_0^1f(t)\, dt\bigg)^4}\, dx\le \sqrt{2}\int_0^1f^2(x)\,dx$$
where $f$ is a function integrable on $[0,1]$ with real values.
My initial thought was that the inequality is trivial if:
$$\int_0^1f(t)\, dt \leq f(x)$$
but this is not true always. Then I thought with Cauchy-Bunyakovsky-Schwarz inequality for integrals:
$$\bigg(\int_0^1f(t)\, dt\bigg)^4 \leq \bigg(\int_0^1f^2(t)\, dt\bigg)^2\leq \int_0^1f^4(t)\, dt$$
but I don't know if this inequality is true:
$$\int_0^1\sqrt{f^4(x)+\int_0^1f^4(t)\, dt}\, dx\le \sqrt{2}\int_0^1f^2(x)\,dx$$
It might be true, but I don't know how to prove it. I would appreciate any help.
Notice that for any two real numbers $a$ and $b$, we have:
$$a^4+b^4=2(a^2-ab+b^2)^2-(a-b)^4\leq 2(a^2-ab+b^2)^2$$
Therefore:
$$\sqrt{a^4+b^4} \leq \sqrt{2}(a^2-ab+b^2)$$
Set $a=f(x)$ and $b=\displaystyle\int_{0}^1f(t)\, dt$ to get that:
$$\sqrt{f^4(x)+\left(\int_0^1f(t)\, dt\right)^4}\, dx\le \sqrt{2}\left[f^2(x)-f(x)\int_{0}^1f(t)\, dt+\left(\int_{0}^1f(t)\, dt\right)^2\right]$$
Now, integrate from $0$ to $1$ with respect to $x$ and notice that:
$$\int_0^1\left(f(x)\int_{0}^1f(t)\, dt\right)\, dx=\int_{0}^1f(t)\, dt\int_{0}^1f(x)\, dx=\left(\int_{0}^1f(t)\, dt\right)^2$$
to arrive at:
$$\int_0^1\sqrt{f^4(x)+\left(\int_0^1f(t)\, dt\right)^4}\, dx\leq \sqrt{2}\int_0^1 f^2(x)\,dx$$