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Proving $a^a \times b^b \gt (\frac{a+b}{2})^{(a+b)}$,where $a \gt b \gt 0$

One method that could be used here is using the inequality $$(1+x)^{(1+x)} \times (1-x)^{(1-x)} \gt 1$$ nothing is wrong with this approach however some examiners might like to see the first proof too,the first proof is also very easy,but I was just wondering whether there is any other way to prove the same?

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up vote 5 down vote accepted

You can use the weighted AM >= GM

$$\left(\frac{2}{a+b}\right)^{a+b} = \left(\frac{ a \times 1/a + b \times 1/b}{a+b}\right)^{a+b} \ge (1/a)^a (1/b)^b$$

Rewriting gives us your inequality.

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$\quad$+1,Short and simple :-) – Quixotic Aug 9 '11 at 14:19

Assuming $a,b>0$ we get

$a^a.b^b\ge \left(\frac{a+b}2\right)^{(a+b)}$ $\Leftrightarrow$

$a\log a+b\log b \ge (a+b)\log\left(\frac{a+b}2\right)$ $\Leftrightarrow$

$\frac{a\log a+b\log b}2 \ge \frac{a+b}2\log\left(\frac{a+b}2\right)$

The last inequality is true if $f(x)=x\log x$ is convex for $x>0$, which is true, since the derivative $f'(x)=\log x+1$ in increasing. (The last inequality has the form $\frac{f(a)+f(b)}2 \ge f\left(\frac{a+b}2\right)$.)

If $a\ne b$ we get strict inequality since $f(x)$ is strictly convex ($f'(x)$ is strictly increasing.)

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Your inequality is equivalent to $$ \left(\frac{a}{a+b}\right)^a \left(\frac{b}{a+b}\right)^b > \left(\frac{1}{2}\right)^{a+b}. $$

Let $p = a/(a+b)$. Then $1/2 < p < 1$ and the inequality is $$ p^a (1-p)^b > (1/2)^{a+b}. $$ Suppose (wlog) that $a,b$ are integers. Define $f(q) = q^a (1-q)^b$, the probability that a coin with bias $q$ comes up $a$ times "head" and $b$ times "tails". The inequality $f(p) > f(1/2)$ can be generalized to $$ \frac{a}{a+b} = \operatorname*{argmax}_{q \in [0,1]} f(q). $$ This states that the Maximum Likelihood estimate for $q$ is $a/(a+b)$.

If you don't like the fact that $a,b$ should be interpreted as integers, you can consider the equivalent inequality $$ g(p) > g(1/2), \quad g(q) = q^p (1-q)^{1-p}. $$ This has the same interpretation. Moreover, the "log likelihood ratio" $$\log \frac{g(p)}{g(1/2)}$$ is exactly equal to the Kullback-Leibler divergence between a $p$-biased coin and a fair coin. Non-negativity of the latter is equivalent to the inequality.

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