If $a$ and $b$ are arbitrary positive real numbers and $m$ an integer, prove that \[\Bigr( 1+\frac ab \Bigl)^m +\Bigr( 1+\frac ba \Bigl)^m \geq 2^{m+1}.\]
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Tags: induction, function, Inequality, algebra, 2-variable inequality, IMO Shortlist, inequalities
-Elixir-
23.09.2010 16:21
Induction on $m.$
Assume WLOG $a \ge b \Rightarrow \frac{a}{b} \ge \frac{b}{a}.$
Case $m = 0$ is trivial.
Then, assume it is true for $m=k.$ By Chebyshev's, we then have
\[\begin{array}{c}\ \left(1 + \frac{a}{b}\right)^{m+1} + \left(1 + \frac{b}{a}\right)^{m+1} \ge \frac{1}{2}\left(2 + \frac{a}{b} + \frac{b}{a}\right)\left(\left(1 + \frac{a}{b}\right)^m + \left(1 + \frac{b}{a}\right)^m\right) \ge 2^{m+2}, \end{array}\]
the last part proven by AM-GM and the inductive hypothesis.
For negative $m$, transform $\frac{a}{a+b} = x, \frac{b}{a+b} = y,$ and thus the problem is now given $x + y = 1,$ prove $x^m + y^m \ge 2^{m-1},$ which can follow from induction and Chebyshev's.
sonhadhsphn
23.09.2010 16:38
amparvardi wrote: If $a$ and $b$ are arbitrary positive real numbers and $m$ an integer, prove that \[\Bigr( 1+\frac ab \Bigl)^m +\Bigr( 1+\frac ba \Bigl)^m \geq 2^{m+1}.\] $(1+\frac{a}{b})(1+\frac{b}{a}) \ge 2\sqrt{\frac{a}{b}}.2\sqrt{\frac{b}{a}}=4$ $(1+\frac{a}{b})^m+(1+\frac{b}{c})^m \ge 2\sqrt{[(1+\frac{a}{b})(1+{\frac{b}{a})}]^m} \ge 2(\sqrt{4})^m =2^{m+1}$
-Elixir-
23.09.2010 16:48
That proof was so much simpler. As a matter of fact I think it implies the inequality for all real $m.$ Edit1: Wait, does that prove it for negative $m$ as well? Edit2: It does. Nevermind.
sayantanchakraborty
29.12.2013 07:03
A nice proof by jensen's inequality We note that the function $f(x)=x^m$ is convex on the intervals $[1,\infty)$ and $[-1,-\infty)$. So applying JENSEN's inequality for m in each of the intervals we get \[(1+a/b)^m+(1+b/a)^m>2(1/2(1+a/b)+1/2(1+b/a))^m=2(1+1/2(a/b+b/a))^m>=2^{m+1}\] For m=0 there is nothing to prove.
opptoinfinity
20.10.2018 18:56
The inequality is equivalent to $\frac{(a+b)^m}{b^m} + \frac{(a+b)^m}{a^m} \ge 2^{m+1}$. Which is again equivalent to $\frac{1}{2}.(\frac{1}{a^m}+\frac{1}{b^m})\ge (\frac{2}{a+b})^m$. Now consider the function function $f(x)=\frac{1}{x^m}$ which is convex $\forall m \in \mathbb{Z}.$ Done by applying Jensen's inequality!