3. Let n > 1 be an integer and $a_1, a_2, . . . , a_n$ be positive reals with sum 1. a) Show that there exists a constant c ≥ 1/2 so that $\sum \frac{a_k}{1+(a_0+a_1+...+a_{k-1})^2}\geq c$, where $a_0 = 0$. b) Show that ’the best’ value of c is at least $\frac{\pi}{4}$.
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Tags: inequalities, calculus, integration
Tintarn
30.10.2017 08:58
For the first one, just note that $a_0+a_1+\dots+a_k \le 1$ and hence $LHS \ge \sum \frac{a_k}{2} \ge \frac{1}{2}$.
For the second one, put $a_k=\frac{1}{n}$ and observe that these are Riemann Sums for $\int_0^1 \frac{1}{1+x^2} dx=\frac{\pi}{4}$.
Also note the equivalent continuous version: For $f:[0,1] \to [0,1]$ increasing and differentiable with $f(0)=0, f(1)=1$, we have $\int_0^1 \frac{f'(x)}{1+f(x)^2} dx \ge \frac{\pi}{4}$. Here the obvious candidate for equality is $f(x)=x$.
whiwho
30.10.2017 17:57
GGPiku wrote: 3. Let n > 1 be an integer and $a_1, a_2, . . . , a_n$ be positive integers with sum 1. a) Show that there exists a constant c ≥ 1/2 so that $\sum \frac{a_k}{1+(a_0+a_1+...+a_{k-1})^2}\geq c$, where $a_0 = 0$. b) Show that ’the best’ value of c is at least $\frac{\pi}{4}$. Quote: $a_1, a_2, . . . , a_n$ be positive integers with sum 1 Doesn't this mean that all $a$'s but one are ceros and that one is a $1$? But a cero isn't a positive integer, so it can't be unless $n=1$, which contradicts the statement. Should it maybe be over positive reals?