<s>[INST] proving 1/x is convex (without differentiating) 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 $\alpha = 0.5$).
 [/INST] We want to show that
$\frac{1}{a x + (1-a) y } 
\leq \frac{a}{x}+\frac{1-a}{y}
$
The right side is
$ \frac{a}{x}+\frac{1-a}{y}
=\frac{ay+(1-a)x}{xy}
=\frac{a(y-x)+x}{xy}
$
and the left side is
$\frac{1}{a x + (1-a) y }
=\frac{1}{a (x-y) + y }
$
so we want
$\frac{1}{a (x-y) + y} 
\le \frac{a(y-x)+x}{xy}
$.
Cross multiplying,
this is
$\begin{array}\\
xy
&\le (a (x-y) + y)(a(y-x)+x)\\
&=-a^2(x-y)^2+ay(y-x)+ax(x-y)+xy\\
&=-a^2(x-y)^2+a(x-y)(x-y)+xy\\
&=-a^2(x-y)^2+a(x-y)^2+xy\\
&=(a-a^2)(x-y)^2+xy\\
&=a(1-a)(x-y)^2+xy\\
\end{array}
$
And since both
$a(1-a) \ge 0$
and
$(x-y)^2 \ge 0$
the result follows.
</s>