I have shown that
$x \log x + (−x) \log( − x) ≥ k$
is a rejection region for the likelihood ratio test for the p parameter of a binomial distribution, where,
$H_0: = 1/2$
$ 1 : ≠ 1/2$
Next, I am asked to show that the above rejection region can be simplified in a way that doesn't involve the log function, using the symmetry property of the LHS function.
I have tried plotting the chart, and indeed it is a symmetric function. Have tried rewriting it into a quadratic function, however failed to complete the square.
Consider the function
$$g(x)=x\ln x+(n-x)\ln(n-x)\quad,\,x\in (0,n)$$
Differentiating, we have
$$g'(x)=\ln x-\ln(n-x)$$
And $$g''(x)=\frac1x + \frac1{n-x}>0 \quad\forall \,x\in (0,n)\,,$$
so that $g$ is convex in $x$.
Also,
$$ g'(x) \begin{cases} >0 &,\text{ if }x>\frac{n}2 \\<0 &,\text{ if }x<\frac{n}2 \\=0 &,\text{ if }x=\frac{n}2 \end{cases} $$
This gives a fair idea about a picture of $g(x)$ versus $x$. The function decreases for $0<x<\frac{n}2$, reaches a minimum at $x=\frac{n}2$, and then increases for $\frac{n}2<x<n$. In other words, a plot of $g(x)$ should give a U-shaped curve.
In your case, $g(x)$ is defined for $x\in \{1,2,\ldots,n-1\}$. But the conclusion above remains same.
Draw a horizontal line that cuts $g(x)$ at two-points $k_1,k_2$ (say) where $k_2>k_1$. This is the $g(x)=k$ line. Now based on the nature of $g(x)$, what can you say about the region $g(x)\ge k $ in terms of $x$?