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I know that the function $x \mapsto \frac1x$ 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 $\alpha = 0.5$).

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    $\begingroup$ Compute the right-hand side minus the left-hand side. Then factor the fraction you get. $\endgroup$
    – David
    Commented Jan 22, 2016 at 7:03
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    $\begingroup$ Since $1/x$ is continuous, inequality for $\alpha=0.5$ implies inequality for all $\alpha\in[0,1]$ (just think of binary search for $\alpha$). $\endgroup$
    – A.S.
    Commented Jan 22, 2016 at 7:48
  • $\begingroup$ Thanks for the comments, @A.S. this is interesting, why this is true, can you put it in a more formal way. $\endgroup$
    – MrX
    Commented Jan 22, 2016 at 7:52
  • $\begingroup$ $\alpha=1/2$ implies all $\alpha$ with finite binary expansion. Continuity extends it to all $\alpha$. Serpinski theorem says that Lebesgue integrability is sufficient for extension. $\endgroup$
    – A.S.
    Commented Jan 22, 2016 at 8:18
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    $\begingroup$ I see, thanks @dxiv , I was not aware that mid point convexity implies convexity for continuous functions although it kind of intuitive.(the theorem by Jake). $\endgroup$
    – MrX
    Commented Jan 24, 2016 at 23:40

2 Answers 2

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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.

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

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  • $\begingroup$ Yup. So many inequalities come down to $x^2 \ge 0$. And to think we did this over 5 years ago. $\endgroup$ Commented Apr 7, 2021 at 14:47

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