We introduce the variables
so that
and all . Simplifying and homogenizing, we need to prove
Consider some . AM-GM tells us that
is at least

\[\begin{align}\sqrt{(j+1)(j+4)}(x_n+\cdots+x_{j+1})\ge(j+2)(x_n+\cdots+x_{j+1}),
\end{align}\]

where
Hence if we can show that
then adding this with (1), we have
Note that when , this inequality is equivalent to the desired inequality. So it suffices to show the inequality for , which is simply

\[\begin{align*}
\frac{x_n^2}{x_{n-1}}&>\frac{1}{4}\sum_{i=n-1}^{n}(i(i+1)-2((n-1)+1)((n-1)+2))x_i\\
&=-\frac{1}{4}n((n+3)x_{n-1}+(n+1)x_n).
\end{align*}\]

The LHS is positive while the RHS is negative, so we're done.

Motivation: We should get rid of one by one in that order, since it's achievable.

We use the substitution
Note that

\[\begin{align*}
\sum\frac{1}{a(1+b)}&=\sum\frac{1}{\frac{ty}{x}\left(1+\frac{tz}{y}\right)}\\
&=\sum\frac{x}{t(y+tz)}\\
&=\sum\frac{x^2}{txy+t^2xz}\\
&\ge\sum\frac{(x+y+z)^2}{(t+t^2)\sum{yz}}\\
&\ge\sum\frac{3\sum{yz}}{(1+t^3)\sum{yz}}\\
&=\frac{3}{1+abc},\]

as desired, with equality iff and , i.e. .