proving 1/x is convex (without differentiating)

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.

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 $$ a/x+b/y\geq 1/(a x+b y)$$ $$\iff (a x+ b y)(a y+b x)\geq x y$$ $$\iff (a^2+b^2)x y +a b(x^2+y^2)\geq x y $$ $$\iff (1-2 a b)x y +a b (x^2+y^2)\geq x y$$ $$ \iff -2 a bx y+a b(x^2+y^2)\geq 0$$ $$ \iff a b (x-y)^2\geq 0.$$