If $a, b>0$ and n is a positive integer, prove that: $(1+\frac{a}{b})^n+(1+\frac{b}{a})^n\ge 2^{n+1}$

Question

If $a, b>0$ and n is a positive integer, prove that: $(1+\frac{a}{b})^n+(1+\frac{b}{a})^n\ge 2^{n+1}$

I solved it in the following way:

$(1+\frac{a}{b})^n+(1+\frac{b}{a})^n \ge (2\sqrt{\frac{a}{b}})^n+(2\sqrt{\frac{b}{a}})^n=2^n((\frac{a}{b})^n+(\frac{b}{a})^n)$

It is enough if $(\sqrt{\frac{a}{b}})^n+(\sqrt{\frac{b}{a}})^n\ge2$

If $n=1$:

$\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\ge2$

$\frac{a}{b}+\frac{b}{a}+2\ge 4$ which holds true.

If it holds true for $n=k$:

If $b\ge a$: $(\sqrt{\frac{a}{b}})^k+(\sqrt{\frac{b}{a}})^k$

$(\sqrt{\frac{a}{b}})^{k+1}+\sqrt{\frac{a}{b}}*(\sqrt{\frac{b}{a}})^k\ge 2$

$(\sqrt{\frac{a}{b}})^{k+1}+(\sqrt{\frac{b}{a}})^{k-1}\ge 2$

And since $b\ge a$ We have that the original proposition holds true for k+1. Similarly, we prove that it is true if $a\ge b$. Hence because of induction, it holds true.

Could you please show some other methods of solving this question easier?

Answer 1

Use Mean power inequality: $$\frac{x^n+y^n}{2} \ge \left(\frac{x+y}{2}\right)^n, n>1$$ Then $$F=\frac{(1+a/b)^n+(1+b/a)^n}{2} \ge \left( \frac{2+a/b+b/a}{2} \right)^n$$ Further by AM-GM: $a/b+b/a \ge 2$, we get $F\ge 2^n$ and hence the result.

Answer 2

Using the classic inequality $x^2 + y^2\geq 2xy$, you have $\frac{a}{b}+\frac{b}{a}\geq 2$ and $$ \left(1+\frac{a}{b}\right)^n+\left(1+\frac{b}{a}\right)^n≥2\left(\sqrt{\left(1+\frac{a}{b}\right)\left(1+\frac{b}{a}\right)}\right)^n = 2\left(\sqrt{ 2+\frac{a}{b}+\frac{b}{a}}\right)^n\geq 2\left(\sqrt{ 2+2}\right)^n = 2^{n+1} $$

Answer 3

Apply AM-GM to get, $(1+a/b)^n+(1+b/a)^n \ge 2*((1+a/b)(1+b/a))^{n/2}$.

hence it suffices to show $(1+a/b)(1+b/a) \ge 4 $.But for that just expand and apply am-gm again

Answer 4

$x = 1 + \frac ab$ and $y = 1 + \frac ba$ are positive and add up to $x + y = 2 + \frac ab + \frac ba \ge 4$.

So we start with $x^n + y^n \ge 2^{n+1}$ when $n=1$.

To induct on $n$, suppose that $x > y$. Then $$ x^{n+1} + y^{n+1} = x \cdot x^n + y \cdot y^n \ge x \cdot x^n + (4-x) \cdot y^n \ge 2 \cdot x^n + 2 \cdot y^n. $$ The first inequality holds because $y \ge 4-x$; the second holds because we're adding $(2-x)(x^n - y^n) < 0$.