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?
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 $t\ge 0$.