Question: Let $f,g:[0,1]\to\mathbb{R}$ be two continuous functions such that $f(x)g(x)\ge 4x^2, \forall x\in[0,1].$ Prove that $$\left|\int_0^1 f(x)dx\right|\ge 1\text{ or }\left|\int_0^1 g(x)dx\right|\ge 1.$$
Solution: Since we have $f(x)g(x)\ge 4x^2\ge 0, \forall x\in[0,1]\implies f(x)g(x)\ge 0, \forall x\in[0,1].$ This in turn implies that $f(x)$ and $g(x)$ has the same sign at each $x\in[0,1]$.
Claim: $f$ does not change it's sign in the interval $[0,1]$.
Proof: Let us assume that $f(x)>0$ at some $x=a$ and $f(x)<0$ at some $x=b$, where $a,b\in[0,1]$ and $b>a$. Now since $f$ is continuous in $[0,1]$, thus by IVT we can conclude that $\exists c\in(a,b),$ such that $f(c)=0$. Hence, we have $f(c)g(c)=0\ge 4c^2\implies 4c^2\le 0,$ but $c>0\implies 4c^2>0.$ Hence we arrive at a clear contradiction. Thus, we can conclude that $f$ does not change it's sign in the interval $[0,1]$.
Using our claim, we can also conclude that, $g$ does not change it's sign in the interval $[0,1]$. Thus, either $f(x),g(x)>0, \forall x\in[0,1]$ or $f(x),g(x)<0, \forall x\in[0,1]$.
Observe that in any case $$\left|\int_0^1 f(x)dx\right|=\int_0^1|f(x)|dx\text{ and }\left|\int_0^1 g(x)dx\right|=\int_0^1|g(x)|dx.$$
Thus, by Cauchy-Schwarz inequality we have $$\left|\int_0^1 f(x)dx\right|\left|\int_0^1 g(x)dx\right|=\int_0^1|f(x)|dx\int_0^1|g(x)|dx\\=\int_0^1\left(\sqrt{|f(x)|}\right)^2dx\int_0^1\left(\sqrt{|g(x)|}\right)^2dx\\\ge \left(\int_0^1\sqrt{|f(x)|}.\sqrt{|g(x)|}dx\right)^2=\left(\int_0^1\sqrt{|f(x)g(x)|}dx\right)^2=\left(\int_0^1\sqrt{f(x)g(x)}dx\right)^2.$$
Now since $\forall x\in[0,1]$, we have $$f(x)g(x)\ge 4x^2\implies \sqrt{f(x)g(x)}\ge 2x\\\implies\int_0^1\sqrt{f(x)g(x)}dx\ge \int_0^1 2x dx=1\\\implies \left(\int_0^1\sqrt{f(x)g(x)}dx\right)^2\ge 1.$$
Hence, we can conclude that $$\left|\int_0^1 f(x)dx\right|\left|\int_0^1 g(x)dx\right|\ge 1.$$
Thus, we can conclude that $$\left|\int_0^1 f(x)dx\right|\ge 1\text{ or }\left|\int_0^1 g(x)dx\right|\ge 1.$$ Hence, we are done.
Is this solution correct and rigorous enough? Are there any alternative solutions?