Key Investment Insights
<|endoftext|>## Mathematical Forums
## Category: High School Olympiads
## Topic: Find the maximum and minimum value
## Views: 106
## [enter: math-user1, num_posts=14, num_likes_received=0]
## [math-user1, num_likes=0]
Assuming that $x$ is a positive real number, find the maximum and minimum value of
$$f(x)=x+\frac{1}{x}+\frac{1}{x+1}$$
## [enter: math-user2, num_posts=701, num_likes_received=447]
## [math-user2, num_likes=0]
Note that
\[f(x)=x+\frac{1}{x}+\frac{1}{x+1}=x+1+\frac{1}{x}+\frac{x}{x+1}-1=x+1+\frac{x^2+1}{x(x+1)}-1=x+1+\frac{(x-1)^2}{x(x+1)}\]
Thus, we have that
\[f(x)\geq 2\sqrt{x\cdot\frac{(x-1)^2}{x(x+1)}}+1=2\sqrt{\frac{(x-1)^2}{x+1}}+1\geq 1\]
Thus, the minimum is $1$, achieved at $x=1$. We also note that
\[f(x)\leq x+1+\frac{x^2+1}{4x}=\frac{4x^2+4x+4}{4x}=\frac{(2x+1)^2+3}{4x}\]
Thus, we have that
\[\frac{(2x+1)^2+3}{4x}\geq 2\]
or
\[(2x+1)^2-8x+3\geq 0\]
or
\[(
Management Rights Multiplier: 5.20
ROI Estimate: 19.24%
