4
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ $\quad$+1,Short and simple :-) $\endgroup$ – Quixotic Aug 9 '11 at 14:19
6
$\begingroup$

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.)

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.