Prove that $\int_0^1\sqrt{f^4(x)+(\int_0^1f(t)\, dt)^4}\, dx\le \sqrt{2}\int_0^1f^2(x)\,dx$

Question

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.

Answer 1

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:

$$\int_0^1\sqrt{f^4(x)+\left(\int_0^1f(t)\, dt\right)^4}\, dx\leq \sqrt{2}\int_0^1 f^2(x)\,dx$$