1.26
The quantities $a_i$ in this exercise are all positive real numbers.
(a) show that
$$ a_1 a_2 \leqslant \left( \frac{a_1 + a_2}{2} \right) ^2 . $$
(b) hence prove, by induction on m, that
$$ a_1 a_2 . . . a_p \leqslant \left( \frac{a_1 + a_2 + . . . + a_p}{p} \right) ^p ,$$
where $p = 2^m$ with $m$ a positive integer. Note that each increase of $m$ by unity doubles the number of factors in the product.
Attempts:
(a)
Have gotten this far with (a) by showing that: $$ \left( \frac{a_1 + a_2}{2} \right) ^2 = a_1 a_2\left( \frac{1}{2} + \frac{a_1}{4 a_2} + \frac{a_2}{4a_1} \right) $$
(b)
inducing from the case for m = M to the case of m = M + 1, is the same as going from p -> 2p
I figure that I'm trying to go from this expression:
$$ \left( \frac{a_1 + a_2 + . . . + a_p}{p} \right) ^p \left( a_{p+1} . . . a_{2p} \right) $$
to show that it contains this expression atleast:
$$ \left( \frac{a_1 + a_2 + . . . + a_{2p}}{2p} \right) ^{2p} $$
with guidance from a comment I figured that I can take the first expression and rewrite the second term using the induction hypothesis to give:
$$ \left( \frac{a_1 + a_2 + . . . + a_{p}}{p} \right) ^{p}\left( \frac{a_{p+1} + a_{p+2} + . . . + a_{2p}}{p} \right) ^{p} $$
(a) This is the AM-GM inequality: $$0\leq(a_1-a_2)^2=(a_1+a_2)^2-4a_1a_2$$
(b) Let $j_1=\left( \frac{a_1 + a_2 + . . . + a_{p}}{p} \right)$, $j_2=\left( \frac{a_{p+1} + a_{p+2} + . . . + a_{2p}}{p} \right)$. Use (a): $$j_1j_2\leq\left( \frac{j_1 + j_2}{2} \right) ^2$$ So $$ \left( \frac{a_1 + a_2 + . . . + a_{p}}{p} \right)^p \left( \frac{a_{p+1} + a_{p+2} + . . . + a_{2p}}{p} \right)^p\leq \left( \frac{a_1 + ...+a_p +a_{p+1}+ . . . + a_{2p}}{2p} \right)^{2p} $$ As I said in the comments, this book is really great, keep it up!