show that $f$ is log concave $\frac{1}{\sum_{i=1}^{n}\frac{1}{x_i}}$

Question

given $f(x)=\frac{1}{\sum_{i=1}^{n}\frac{1}{x_i}}$ over $\mathbb{R}^n_{++}$ show that that $f$ is log-concave.

I've got to $-\ln(\sum_{i=1}^{n}\frac{1}{x_i})$ and tried using the definition but didn't go well so I thought using maybe the hessian criterion but got complex aswell
any help?

Answer 1

Let's do the 2d case as a hint:

So we want to show:

$$ln (⁡ f ( θ x + ( 1 − θ ) y )) ≥ θ ln ⁡ f ( x ) + ( 1 − θ ) ln ⁡ f ( y ) $$

for $0<\theta <1$.

The left hand side is:

$$-ln( \frac{1}{θ x_{1} + ( 1 − θ ) y_{1}} + \frac{1}{θ x_{2} + ( 1 − θ ) y_{2}})$$

The right hand side is:

$$-\theta ln(\frac{1}{x_{1}}+\frac{1}{x_{2}}) - (1-\theta) ln(\frac{1}{y_{1}}+\frac{1}{y_{2}})$$

Getting common denominators and simplifying the left hand side is:

$$ln(θ x_{1} + ( 1 − θ ) y_{1}) + ln(θ x_{2} + ( 1 − θ ) y_{2}) - ln(θ x_{1} + ( 1 − θ ) y_{1} + θ x_{2} + ( 1 − θ ) y_{2} )$$

And the right hand side is:

$$\theta ln(x_{1}) + \theta ln(x_{2}) - \theta ln (x_{1}+x_{2}) + \gamma ln(y_{1}) + \gamma ln(y_{2}) - \gamma ln (y_{1}+y_{2})$$

where $\gamma = 1 - \theta$.

Now simplifying we get:

$$(x_{1}+x_{2})^{\theta}(y_{1}+y_{2})^{\gamma}(θ x_{1} + ( 1 − θ ) y_{1})(θ x_{2} + ( 1 − θ ) y_{2}) ≥ (x_{1}x_{2})^{\theta}(y_{1}y_{2})^{1-\theta}( θ x_{1} + ( 1 − θ ) y_{1} + θ x_{2} + ( 1 − θ ) y_{2} )$$