# Solution to AMM problem 11855

$\textbf{Problem}$
$\hspace{2cm}$ For a continuous and nonnegative function f on $[0, 1]$, let $\mu_{n}=\int_0^1 {x^{n}f(x)}dx$. $\textbf{Prove that:}$
$\mu_{n+1} \mu_{0} \geq \mu_{n} \mu_{1}$.
$\textbf{Solution}$
Let f be continuous, nonnegative function and g be continuous, positive function on [0, 1] and $n\in \mathbb{N}$. Show that
$[\int_0^1 {g^{n+1}(x)f(x)}dx][\int_0^1 {f(x)}dx] \geq [\int_0^1 {g^{n}(x)f(x)}dx][\int_0^1 {g(x)f(x)}dx]$ $\hspace{2cm} (1)$
Proof:
If $f(x)=0$ for all $x\in [0, 1]$, the inequality (1) is hold.
If $f(x)\neq 0$ that mean there is $x_{0}\in [0, 1]$ such that $f(x_{0})$>0(note that f is nonnegative). On the other hand, one has f, g are continuous on [0, 1] and g is positive so $\int_0^1 {g^{n}(x)f(x)}dx$>0 for all $n\in \mathbb{N}$ and $\int_0^1 {f(x)}dx$>0.
Then the inequality is equivalent to:
$\displaystyle \frac{\int_0^1 {g^{n+1}(x)f(x)}dx}{\int_0^1 {g^{n}(x)f(x)}dx} \geq \frac{\int_0^1 {g(x)f(x)}dx}{\int_0^1 {f(x)}dx}$
Let $a_{n}=\frac{\int_0^1 {g^{n}(x)f(x)}dx}{\int_0^1 {g^{n-1}(x)f(x)}dx}$, we will prove that $(a_{n})_{n\in N, n\geq 1}$ is increasing,
That mean
$\displaystyle a_{n+1}\geq a_{n} \Longleftrightarrow \frac{\int_0^1 {g^{n+1}(x)f(x)}dx}{\int_0^1 {g^{n}(x)f(x)}dx} \geq \frac{\int_0^1 {g^{n}(x)f(x)}dx}{\int_0^1 {g^{n-1}(x)f(x)}dx} \Longleftrightarrow [\int_0^1 {g^{n+1}(x)f(x)}dx][\int_0^1 {g^{n-1}(x)f(x)}dx] \geq [\int_0^1 {g^{n}(x)f(x)}dx]^{2}$
It is easy to see it, by the inequality Cauchy-Schwarz for integral:
One has
$[\int_0^1 {g^{n}(x)f(x)}dx]^{2}=[\int_0^1 {g^{\frac{n-1}{2}}(x)f^{\frac{1}{2}}(x)}dx{g^{\frac{n+1}{2}}(x)f^{\frac{1}{2}}(x)}dx]^{2} \leq [\int_0^1 {g^{n+1}(x)f(x)}dx][\int_0^1 {g^{n-1}(x)f(x)}dx]$.
(It is our claim.)
From the fact $(a_{n})_{n\in N,n\geq 1}$ is increasing, we get:
$a_{n+1} \geq a_{1}$, that mean:
$[\int_0^1 {g^{n+1}(x)f(x)}dx][\int_0^1 {f(x)}dx] \geq [\int_0^1 {g^{n}(x)f(x)}dx][\int_0^1 {g(x)f(x)}dx]$ (we have done)
Now the problem of C.lupu, we only need to choose g(x)=x (One has x>0 for all $\in (0, 1]$-that is sufficent. And g(x)=x is obviously continuous on [0, 1])