An Inequality with Just Two Variable And an Integer

From means the inequality

(1)

$\displaystyle\frac{a}{b^n}+\frac{b}{a^n}\geq 2\sqrt{\frac{ab}{a^nb^n}}.$

(2)

$\displaystyle\frac{a^n}{b^n}+\frac{b}{a}\geq 2\sqrt{\frac{a^nb}{b^na}}.$

(3)

$\displaystyle\frac{b^n}{a^n}+\frac{a}{b}\geq 2\sqrt{\frac{b^na}{a^nb}}.$

By multiplying the relationship (1),(2),(3):

(4)

$\displaystyle\Bigr(\frac{a}{b^n}+\frac{b}{a^n}\Bigr)\Bigr(\frac{a^n}{b^n}+\frac{b}{a}\Bigr)\Bigr(\frac{b^n}{a^n}+\frac{a}{b}\Bigr)\geq \frac{8}{\sqrt{a^{n-1}\cdot b^{n-1}}}.$

We prove that

(5)

$\displaystyle\frac{a^n}{b}+\frac{b^n}{a}\geq a^{n-1}+b^{n-1}.$

$\begin{align} &a^{n+1}+b^{n+1}\geq a^nb+ab^n\\ &a^n(a-b)-b^n(a-b)\geq 0\\ &(a-b)(a^n-b^n)\geq 0\\ &(a-b)^2(a^{n-1}+a^{n-2}b+\ldots+b^{n-1})\geq 0, \end{align}$

which is true. Now multiply the relationships (4),(5)

$\displaystyle\begin{align} &\Bigr(\frac{a^n}{b}+\frac{b^n}{a}\Bigr)\Bigr(\frac{a}{b^n}+\frac{b}{a^n}\Bigr)\Bigr(\frac{a^n}{b^n}+\frac{b}{a}\Bigr)\Bigr(\frac{b^n}{a^n}+\frac{a}{b}\Bigr)\geq \frac{8(a^{n-1}+b^{n-1})}{\sqrt{a^{n-1}\cdot b^{n-1}}}\\ &\qquad\qquad=8\Biggl(\sqrt{(\frac{a}{b})^{n-1}}+\sqrt{(\frac{b}{a})^{n-1}}\Biggl) \end{align}$

Note that

$\displaystyle\left(\frac{1}{2} \left(\left(\frac{b}{a}\right)^{n-1}+\left(\frac{a}{b}\right)^{n-1}\right)\right)^{\frac{1}{n-1}}\geq \left(\frac{1}{2} \left(\left(\frac{b}{a}\right)^{\frac{n-1}{2}}+\left(\frac{a}{b}\right)^{\frac{n-1}{2}}\right)\right)^{\frac{2}{n-1}},$

By the AM-GM inequality, we can get an $8$ from any combination of $3$ factors from the lhs. By tinkering we get the first, third and fourth terms:

$\displaystyle\begin{align} lhs\, &\geq 2^3 \sqrt{\frac{a^n b^n}{a b}}\left(\frac{a}{b^n}+\frac{b}{a^n}\right)\sqrt{\frac{b a^n}{a b^n}}\sqrt{\frac{a b^n}{b a^n}}\\ &=8 \left(\left(\frac{b}{a}\right)^{\frac{n-1}{2}}+\left(\frac{a}{b}\right)^{\frac{n-1}{2}}\right)\\ &\geq rhs. \end{align}$

This problem, along with a solution (Solution 1), was kindly communicated to me by Dan Sitaru. Dan has eatlier published the problem at the Romanian Mathematical Magazine. Solution 2 is by N. N. Taleb.