Please show that $f(\beta_0,\beta_1)=\log(1+\operatorname{exp}(-y_1(\beta_0+\beta_1 x_1)))+\log(1+\operatorname{exp}(-y_2(\beta_0+\beta_1 x_2)))$ I would like to show that the following result is indeed true. I am very new with this subject, so I ask for a hint to get me started please.

Please show that $f(\beta_0,\beta_1)=\log(1+\operatorname{exp}(-y_1(\beta_0+\beta_1 x_1)))+\log(1+\operatorname{exp}(-y_2(\beta_0+\beta_1 x_2)))$

where $(x_1,y_1), (x_2,y_2)$ are any given data and is convex in $(\beta_0,\beta_1)$
The formula for the logistic regression is given as $$\beta_0+\beta_1 x+\beta_2 x= \log \left(\frac{p}{1-p}\right)$$, where $p$ is the probability. I know that I can check if something is indeed convex being looking at the Hessian matrix.
I recognize that $\frac{p}{1-p}$ is the odds is this is helpful?
The "action" of $f$ is not defined either I don't think, so I'm not sure how to start simplifying the left hand side.
Very vague... I'm not sure what it is a probability of? Hoping someone is familiar with "logistic regression"
Many thanks.

Proof for $f(\beta_0,\beta_1)$ being convex with respect to $(\beta_0,\beta_1)$:
1) $g(x)$ is convex $\Longleftrightarrow  g(a^Tu)$ is convex with respect to $u$ for column matrices $a$ and $u$.
2) Sum of convex functions is convex.
3) $\ln(1+e^x)$ is convex.
Let $g(x) = \ln(1+e^x)$ in 1). Let $a_i=-[y_i\ x_iy_i]^T$. $f(\beta_0,\beta_1)$ is then the sum of convex functions $g(a_i^Tu)$ where $u=[\beta_0\ \beta_1]^T$ which is then convex which is the desired result.
