TITLE: proving 1/x is convex (without differentiating)
QUESTION [12 upvotes]: I know that $\frac{1}{x}$ is convex when  $x \in (0,\infty)$, this can be proven easily by showing that the second derivative is positive. However, I am finding difficulty showing it using the definition of convexity, in other words, for $\alpha \in [0,1]$ and $x_1, x_2 \in \Bbb R^+$, show that:
$$\frac{1}{\alpha x_1 + (1-\alpha) x_2 } \leq \frac{\alpha}{x_1}+\frac{1-\alpha}{x_2}$$
Note that the relation between the harmonic mean and arithmetic mean is just a special case, (take $\alpha = 0.5$).

REPLY [7 votes]: Let $x,y$ be positive  and $a+b=1$ with $a,b\in [0,1].$ We will use the fact that $$a^2+b^2=(a+b)^2-2 a b=1-2 a b.$$  Eliminating the denominators, we have $$ a/x+b/y\geq 1/(a x+b y)$$   $$\iff (a x+ b y)(a y+b x)\geq x y$$ $$\iff  (a^2+b^2)x y +a b(x^2+y^2)\geq x y $$ $$\iff (1-2 a b)x y +a b (x^2+y^2)\geq x y$$  $$ \iff -2 a bx y+a b(x^2+y^2)\geq 0$$ $$ \iff a b (x-y)^2\geq 0.$$