# How to prove a Triangle Inequality

Let $$f:\,\mathbb{R}^2\rightarrow\mathbb{R}_+$$ defined via $$f(x) = \frac{\|x\|^2}{w\cdot x}$$, where $$w\in\mathbb{R}^2$$ is fixed and the dot denotes the inner product. I want to show (or disprove) that, for $$x_1,x_2\in\mathbb{R}^2$$ with $$w\cdot x_1>0$$ and $$w\cdot x_2>0$$, the triangle inequality holds: \begin{align*} f(x_1+x_2)\leq f(x_1)+f(x_2). \end{align*}

Unfornately, just using Cauchy Schwartz and the triangle inequality of the norm doesn't do the trick, so any ideas are welcome.

Note that for positive $$a$$, we have $$a\| x_1\|^2 + \frac{1}{a}\|x_2 \|^2 \ge 2 \|x_1\| \| x_2\| \ge 2x_1\cdot x_2$$ Putting $$a = \frac{w \cdot x_2}{w\cdot x_1}$$, we get $$(w \cdot x_2)f(x_1) + (w\cdot x_1)f(x_2) \ge 2x_1\cdot x_2 \tag{1}$$ Now note that $$w\cdot (x_1+x_2)f(x_1+x_2) = \|x_1 +x_2 \|^2 =\| x_1\|^2 + \|x_2 \|^2 +2x_1\cdot x_2$$ $$= (w\cdot x_1)f(x_1) + (w\cdot x_2)f(x_2) + 2x_1 \cdot x_2 \tag{2}$$ From $$(2)$$ and $$(1)$$, we have $$(w\cdot (x_1+x_2))f(x_1+x_2) \le (w\cdot x_1)f(x_1) + (w\cdot x_2)f(x_2) + (w \cdot x_2)f(x_1) + (w\cdot x_1)f(x_2) = \ (w\cdot(x_1+x_2))(f(x_1) +f(x_2))$$ It follows that $$f(x_1+x_2) \le f(x_1) +f(x_2)$$