Day 19: A Game of Inequality

We have the expression: $(\sum a^3 )(\sum\frac1a )(\sum\frac{1}{a^2})$ Now we can apply the Generalised Hölder's Inequality (for a quick outline of the proof you can consult my unfinished book in the 'A Very Useful Lemma' part').

This yields that the expression is greater than or equal to $\boxed{64}$ ; for example this occurs when all the variables are the same; which is our answer as required.

Applying AM-GM inequality to each of the three terms in the final expression $(\sum a^3 )(\sum\frac1a )(\sum\frac{1}{a^2})$ , we obtain the minimum attainable value as 64.

The inequality is achieved when all three terms equate, or a=b=c=d=1.

Since the link to Michael Ng's book is not working, I'll post the solution he most likely has.

Applying H\"{o}lder's Inequality, in this case, with $\lambda_1=\lambda_2=\lambda_3=\frac{1}{3}$ gives

$(a^3+b^3+c^3+d^3)^\frac{1}{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})^\frac{1}{3}(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{d^2})^\frac{1}{3}$

$\leq (a^3)^\frac{1}{3}\Big(\frac{1}{a}\Big)^\frac{1}{3}\Big(\frac{1}{a^2}\Big)^\frac{1}{3}+(b^3)^\frac{1}{3}\Big(\frac{1}{b}\Big)^\frac{1}{3}\Big(\frac{1}{b^2}\Big)^\frac{1}{3}+(c^3)^\frac{1}{3}\Big(\frac{1}{c}\Big)^\frac{1}{3}\Big(\frac{1}{c^2}\Big)^\frac{1}{3}+(d^3)^\frac{1}{3}\Big(\frac{1}{d}\Big)^\frac{1}{3}\Big(\frac{1}{d^2}\Big)^\frac{1}{3}$

$=1+1+1+1=4$

Cubing both sides of the inequality leads to the result (noting equality occurs when $a=b=c=d=1$ ).