Is it true that $f>g>0 \,\,\Longrightarrow\,\,\int f^2\int g\ge \int f\int g^2 $? Let $f,g: [0,1]\to\mathbb R$, continuous and satisfying
$$
f(x)>g(x)>0, \quad \text{for all $x\in[0,1]$}.
$$
Can we then conclude that 
$$
\frac{\int_0^1f^2dx}{\int_0^1fdx}\geq\frac{\int_0^1g^2dx}{\int_0^1gdx}?
$$
I have no clue how to solve this.
 A: The answer is no. 

In the following by LHS I'll refer to $\int f^2/\int f$, and by RHS $\int g^2/\int g$.
Let's first consider the case where continuity is not demanded. In this case, consider $$f = \begin{cases} 1 & x < 1/2 \\ 1/2 & x \ge 1/2 \end{cases}, \\ \,\,\,\,g = \begin{cases} 1-\varepsilon & x < 1/2 \\ \varepsilon & x \ge 1/2\end{cases},$$ where $\varepsilon \in (0, 1/2)$ is a small number we will choose later. In this case, we get a LHS of $$ \frac{1 + 1/4}{1 + 1/2} = 5/6, $$ and a RHS of $(1-\varepsilon)^2 + \varepsilon^2 = 1 - 2(1-\varepsilon)\varepsilon > 1 - 2\varepsilon$, which is close to $1$ as $\varepsilon \to 0$. It ollows that counterexamples exist.

But we need continuity. Since there's plenty of room between $1 - 2\varepsilon$ and $5/6$, we can attempt a simple fix - linear interpolation. Let $\delta \in (0,1/2)$ be small (we will choose it later). Consider $$ f(x) = \begin{cases} 1 & x < 1/2 - \delta \\ 1/2 & x > 1/2 + \delta \\ \frac{3}{4} - \frac{(x - 1/2)}{4\delta} & x \in [1/2 - \delta, 1/2 + \delta]\end{cases},$$ i.e., $f$ linearly interpolates $1$ and $1/2$ over a window of $2\delta$ in the middle. Do the same thing to $g$. $g< f$ continues to hold.
The calculations are straightforward, so I'll omit them. \begin{align}  
\int f &=   \frac{3}{4} - \delta \\
\int g &= \frac{1}{2} - 2\varepsilon\delta < 1/2 \\ 
\int f^2 &= \frac{5}{8} - \frac{\delta}{12} < 5/8\\ \int g^2 &> (1-2\varepsilon(1-\varepsilon))(1/2 - \delta).
\end{align}
So, \begin{align} \mathrm{LHS} &< \frac{5/8}{3/4 - \delta} = \frac{5}{6} \cdot \frac{1}{1 - 4\delta/3} \\
\mathrm{RHS} &> (1- 2\varepsilon(1-\varepsilon))(1-2\delta) > 1 - 2\delta - 2\varepsilon \end{align}
As we send $\varepsilon$ and $\delta$ to $0$, the RHS is bigger than the LHS, so continuous counterexamples exist. E.g., set $\varepsilon = \delta = 0.01$.
