I am trying to find a proof for this log inequality I found in an Information Theory. The inequality is $$-(a+b)\log(a+b)\leq-a\log a-b\log b\leq-(a+b)\log\frac{a+b}{2}$$ for $a,b\geq0$ and $a+b>0$.
The first inequality seems quite clear, by the fact that $\log$ is an increasing function, but I am struggling to see how to prove the second inequality. Any hints or guidance would be appreciated. Thank you!
On
Recall the weighted AM-GM-HM inequality: for any finite list of non-negative reals $x_i$ and $w_i,$ with $\sum w_i >0,$ $$ \frac{ \sum w_i x_i}{\sum w_i } \ge \left( \prod {x_i}^{w_i}\right)^{1/\sum w_i} \ge \frac{\sum w_i}{ \sum \frac{w_i}{x_i}}.$$
We'll apply this with $w_1 = x_1 = a, w_2 = x_2 = b$. This gives $$ \frac{a^2 + b^2}{a+b} \ge \left( a^a b^b \right)^{1/a+b} \ge \frac{a+b}{ a/a + b/b} = \frac{a+b}{2}.$$
Since $-\log$ is a decreasing function, this gives the inequality (after raising by $a+b$, taking $-\log$, and rearranging, and using the convention $0\log 0 = 0$,)
$$ -(a+b)\log \frac{a^2 + b^2}{a+b} \le - a\log a - b\log b \le -(a+b) \log \frac{a+b}{2}.$$
Finally, note that $ (a+ b)^2 \ge a^2+b^2$ for $a , b \ge 0,$ which means that $-\log(a+b) \le -\log \frac{a^2 + b^2}{a+b}$, finishing the argument.
Rewriting your inequality in the form $$\log((a+b)^{a+b})\geq \log (a^a b^b)$$ let$$a=bt$$ then we get after a rfew steps of algebra: $$(t+1)^{t+1}\geq t^t$$ this is $$(t+1)\ln(t+1)\geq t\ln(t)$$ $$h(t)=(1+t)\ln(1+t)-t\ln(t)$$ $$h'(t)=ln(1+t)-\ln(t)=\ln(\frac{1+t}{t})$$ $$h'(t)=\ln(1+\frac{1}{t})>0$$ for $t>0$ and $$\lim_{t\to 0^+}(1+t)\ln(1+t)-t\ln(t))=0$$