Absolute value inequalities involving multiple variables (Polish Mathematical Olympiad 2016/17) This question is problem 4 from the Polish Mathematical Olympiad 2016/17 Round 1 problem set.

Let $a,b\in\mathbb{R}$ and $0<t<1$. Show that
  $$ |a+(1+t)b| + |a+(1-t)b| \ge \frac{2t}{(2+t)}(|a|+|b|). $$

So I'm trying to solve this inequality and I'm stuck. I've tried to use triangle inequality but that doesn't just work. Is there anyway to solve it in an elegant way?
 A: Using the inequality $|x|\ge x$ and $|x|\ge -x$ for all real $x$, by choosing the signs in each of the two absolute values in the LHS, we obtain the following four inequalities:
\begin{align}|a+(1+t)b|+|a+(1-t)b| &\ge 2(a+b) \\
|a+(1+t)b|+|a+(1-t)b| &\ge 2tb \\
|a+(1+t)b|+|a+(1-t)b| &\ge -2tb \\
|a+(1+t)b|+|a+(1-t)b| &\ge -2(a+b).
\end{align}
We thus have
$$ |a+(1+t)b|+|a+(1-t)b|\ge 2\max(|a+b|,t|b|). $$
It suffices to show that
$$ \max(|a+b|,t|b|)\ge \frac{t}{2+t}(|a|+|b|). $$
First, note that for any $x,y\in\mathbb{R}$ and $0\le\lambda\le 1$, we have $\max(x,y) = \lambda\max(x,y) + (1-\lambda)\max(x,y) \ge \lambda x + (1-\lambda)y$. Taking $x = |a+b|$, $y = t|b|$, and $\lambda = \frac{t}{2+t}$, we have
\begin{align} \max(|a+b|,t|b|)&\ge\frac{t}{2+t}|a+b|+\frac{2}{2+t}t|b|\\
& = \frac{t}{2+t}(|a+b|+|b|+|b|) \\
&\ge \frac{t}{2+t}(|a|+|b|)
\end{align}
where $|a+b|+|b|\ge|a|$ from the triangle inequality.

Note that the above argument also works for any $t\ge 0$, so the result actually is true for all $t\ge 0$.
