Proving $\frac {2n}{(a+b)^n} \le \frac {1}{a^n} + \frac {1}{b^n}$ for $a,b>0, n\in\mathbb{N}$ by induction prove using induction:
$$ \frac {2n}{(a+b)^n} \le \frac {1}{a^n} + \frac {1}{b^n} $$
$$a,b \gt 0  ,  n \in N$$
my attempt:

*

*base $n=1$:
$$ \frac {2}{(a+b)} \le \frac {1}{a} + \frac {1}{b}$$
$$2ab \le b(a+b) + a(a+b)$$
$$2ab \le ab + b^2 + a^2 + ab$$
$$0 <= a^2 + b^2$$
sum of $2$ not negatives, done.


*assume it's right for $n=k \in N$.......

$$ \frac {2k}{(a+b)^k} \le \frac {1}{a^k} + \frac {1}{b^k}$$


*prove for $n=k+1$......

$$ \frac {2(k+1)}{(a+b)^{k+1}} \le \frac {1}{a^{k+1}} + \frac {1}{b^{k+1}}$$
i've tried various things... and get stuck everytime.
this is the attempt mimicking what the lecturer/teaching aid do usually with such problems (problems without parameters tho)...
$$ \frac {2(k+1)}{(a+b)^{k+1}} = $$
$$ \frac {1}{(a+b)} \times \left( \frac {2k}{(a+b)^k} + \frac {2}{(a+b)^k} \right) = $$
$$\left(M + \frac {2}{(a+b)^k} \right) \times \frac {1}{(a+b)}$$
go back to the assumption...
$$M \le \frac {1}{a^k} + \frac {1}{b^k}$$
$$M + \frac {2}{(a+b)^k} \le \frac {1}{a^k} + \frac {1}{b^k} + \frac {2} {(a+b)^k}$$
$$\left( M + \frac {2}{(a+b)^k} \right) \times \frac {1}{a+b} \le \left( \frac {1}{a^k} + \frac {1}{b^k} + \frac {2} {(a+b)^k} \right) \times \frac {1}{a+b}$$
now we need to prove...
$$\left( \frac {1}{a^k} + \frac {1}{b^k} + \frac {2} {(a+b)^k} \right) \times\frac {1}{a+b} \le \frac {1}{a^{k+1}} + \frac {1}{b^{k+1}}$$
no matter how i try messing around with this... i reach a dead end.
please help.

Direct Proof
First, I would re-write the inequality so that you want to show $2n\leq \frac{(a+b)^{n}}{a^n}+\frac{(a+b)^{n}}{b^n}$.  Then, we will use two facts which I will prove before finishing.
Fact 1: Since $a,b>0$ it follows that $(a+b)^{n}=\sum_{i=0}^{n}\binom{n}{i}a^{i}b^{i}\geq \binom{n}{1}ab^{n-1}+\binom{n}{n-1}a^{n-1}b=n(ab^{n-1}+ba^{n-1})$.  (Here I've used the Binomial theorem and just left off some positive terms, so the sum was made smaller.  I could put strict inequality if I like.)
Fact 2: Since $a,b>0$ it follows that $\left(\frac{a}{b}+\frac{b}{a}\right)\geq 2$.  This can be seen as follows:
\begin{align*}
0 &\leq (b-a)^{2} = b^2 - 2ab + a^2 \\
\Rightarrow 2ab &\leq a^{2}+b^{2} \\
\Rightarrow 2   &\leq \frac{a}{b} + \frac{b}{a}
\end{align*}
Now, putting these two facts to use:
\begin{align*}
\frac{(a+b)^{n}}{a^{n}} + \frac{(a+b)^{n}}{b^{n}} &\geq \frac{n(ab^{n-1}+ba^{n-1})}{a^{n}} + \frac{n(ab^{n-1}+ba^{n-1})}{b^{n}} \\
&\geq n\left(\frac{ba^{n-1}}{a^{n}}+\frac{ab^{n-1}}{b^{n}}\right) \\
&=n\left(\frac{b}{a}+\frac{a}{b}\right) \\
&\geq 2n.
\end{align*}
Inductive Proof
Base Case: (sure)
The key fact that we will need is that $(a+b)^{n}\geq a^{n}$ and $(a+b)^{n}\geq b^{n}$.  This is only true when $a,b>0$.
Inductive Step: We know that $2n\leq (a+b)^{n}\left(\frac{1}{a^{n}}+\frac{1}{b^{n}}\right)$ and we want to show that $2n+2=2(n+1)\leq (a+b)^{n+1}\left(\frac{1}{a^{n+1}}+\frac{1}{b^{n+1}}\right)$.
The first thing (as you rightly did) is that we need to make our inductive hypothesis useful by pulling out something that looks like it.
\begin{align*}
(a+b)^{n+1}\left(\frac{1}{a^{n+1}}+\frac{1}{b^{n+1}}\right) &=(a+b)^{n}\left(\frac{a+b}{a^{n+1}}+\frac{a+b}{b^{n+1}}\right) \\
&=(a+b)^{n}\left(\frac{1}{a^{n}}+\frac{1}{b^{n}}\right) + (a+b)^{n}\left(\frac{b}{a^{n+1}}+\frac{a}{b^{n+1}}\right) \\
&\geq 2n + (a^{n}+b^{n})\left(\frac{b}{a^{n+1}}+\frac{a}{b^{n+1}}\right) \\
&\geq 2n + \frac{a^{n}b}{a^{n+1}} + \frac{ab^{n}}{b^{n+1}} \text{ (selective distribution)} \\
&= 2n + \frac{b}{a}+\frac{a}{b} \\
&\geq 2n + 2 \\
&=2(n+1).
\end{align*}
Here, we again used the fact that $\frac{a}{b}+\frac{b}{a}\geq 2$ that I referenced in the other solution.  If it wasn't clear, the selective distribution means that I dropped the middle terms of my "FOIL"ing... i.e. I multiplied only the first two terms together and the last two terms together; the result is smaller because I dropped the middle terms that were positive.
