A proof question of an inequality concerning information theory

Question

I met an inequality proof problem when doing an information theory question. For the convenience of people who want to try to proof it directly, please allow me to write a simple version first.
The question is, proof the following inequality: $$q^2(3-2q)\leq{(q^{1-s}\;{(1-q)}^s+q^s\;{(1-q)}^{1-s})}^3$$ where $0\leq q<\frac12$. Your can use this theorem: for any $0\leq x\leq y\leq1,\;0<s<1,\;x\leq x^{1-s}y^s$ holds.


Thanks to Lee, we can solve the question by using AM-GM inequality, but I got stuck again for the final, developed version of this question.
Firstly, please allow me briefly talk about the background of it.

The original question concerns an error rate calculation by using the repeat code, to say it simply, that is, like we send three '$000$' to denote '$0$' at the sender's side, even one bit got wrong during transmission, the receiver could still make a judgment that the original symbol is '$0$' (like when receiving $001,010,100$)
However, as the communication channel has a wrong probability $q,0\leq q<\frac12$ when sending every symbol, even we use the 3-bit repeat code, we still have a probability of $$\binom32\left(1-p\right)^1+\binom33p^3=q^2(3-2q)$$ to make a wrong judgment due to the situation that more than 2 bits are wrongly transmitted. That is also the source of the first question. And now, if we use N-bit repeat code, please prove $$P_{e,N}\leq\left(2\sqrt{q\left(1-q\right)}\right)^N$$ Yesterday I was quite sparked by Lee's method, but this time I got stuck on the left side, in fact, $P_{e,N}$ is $$P_{e,N}=\sum_{k=\lceil\frac N2\rceil}^N\binom Nkq^k\left(1-q\right)^{N-k}$$ but I have no idea how to sum them up this time... Could you give me some hints? Thank you very much!

Answer 1

Hi and sorry for the very late reply---not sure if you will see this but here is an approach to proving the inequality \begin{equation} q^2(3-2q)\leq{(q^{1-s}\;{(1-q)}^s+q^s\;{(1-q)}^{1-s})}^3 \end{equation} without the need to use the hint you provided.

By observation one can see that the dependency on $s$ is only on the right hand side, and consequently we can opt to optimize over $s$ on the right hand side to find its minimum value first. To do this, an application of the AM-GM inequality does the trick: $$ \begin{split} q^{1-s}\;{(1-q)}^s+q^s\;{(1-q)}^{1-s} &\geq 2\sqrt{q^{1-s} \cdot (1-q)^s \cdot q^{s}\cdot (1-q)^{1-s}} \\ &= 2\sqrt{q(1-q)}. \end{split} $$ The equality holds when $s=\frac{1}{2}$. It then suffices to show that $$ q^2(3-2q)\leq (2\sqrt{q(1-q)})^3 $$ In the case where $q=0$ the above is correct. In the case where $0<q<\frac{1}{2}$, the above is equivalent to showing $$ 64(1-q)^3 \geq q(3-2q)^2. $$ Again, one can notice that over the range of $q$, the left hand side is decreasing in $q$ while the right hand side is increasing in $q$. This implies that $$ 64(1-q)^3 \geq 64\left(1-\frac{1}{2}\right)^3 > \frac{1}{2}\cdot\left(3-2\cdot\frac{1}{2}\right)^2 \geq q(3-2q)^2. $$ Combining the above discussion completes the proof.

(Note: I didn't go through the story to check if the inequality is the right one for the problem; I only proved that the inequality is true.)