AMM problems

\bigstar \textbf{PROBLEM 11855}
\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}.
We will show more general statement:
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)
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])
\bigstar \textbf{PROBLEM 11892}
Let f be a real-valued continuously differentiable function on $[a, b]$ with positive derivative on $(a, b)$. Prove that, for all pairs (x_{1}, x_{2}) with a\leq x_{1} < x_{2} \leq b and f(x_{1})f(x_{2})>0, there exists t\in (x_{1}, x_{2}) such that
\displaystyle \frac{x_{1}f(x_{2})-x_{2}f(x_{1})}{f(x_{2})-f(x_{1})}=t-\frac{f(t)}{f'(t)}
Because f(x_{1})f(x_{2})>0, f(x_{1}) and f(x_{2}) are the same sign. Without lost generality, we can assume that f(x_{1}) and f(x_{2}) are positive (Otherwise, we consider -f). Then for all x\in [x_{1}, x_{2}], by Lagrange’s Theorem we get f(x)-f(x_{1})=f'(c(x))(x-x_{1}) >0, hence f(x)>f(x_{1})>0.
Now, consider two function: g(x)=\frac{x}{f(x)} and h(x)=\frac{1}{f(x)} are continuously differentiable on [x_{1}, x_{2}], by Cauchy’s Theorem there is t\in (x_{1}, x_{2})
\displaystyle \frac{\frac{x_{2}}{f(x_{2})}-\frac{x_{1}}{f(x_{1})}}{\frac{1}{f(x_{2})}-\frac{1}{f(x_{1})}}=\frac{\frac{f(t)-tf'(t)}{f^{2}(t)}}{-\frac{f'(t)}{f^{2}(t)}} \Longleftrightarrow \frac{x_{1}f(x_{2})-x_{2}f(x_{1})}{f(x_{2})-f(x_{1})}=t-\frac{f(t)}{f'(t)}.