Applications of Derivatives
Functional Inequality — Deducing f'
DAILY_CHALLENGE
Grade 12
Question:
Let for a differentiable function $f:(0,\infty)\to\mathbb{R}$, $f(x)-f(y)\geq\log_e\left(\dfrac{x}{y}\right)+x-y$, $\forall x,y\in(0,\infty)$. Then $\displaystyle\sum_{n=1}^{20}f'\left(\dfrac{1}{n^2}\right)$ is equal to
Step-by-Step Solution
Key Concept: From the functional inequality, by letting $y\to x$: $f'(x)\geq\frac{1}{x}+1$. By symmetry (swapping $x$ and $y$), $f'(x)\leq\frac{1}{x}+1$. Hence $f'(x)=\frac{1}{x}+1$.
$f'(x)=\frac{1}{x}+1$. $f'(1/n^2)=n^2+1$. $\sum_{n=1}^{20}(n^2+1)=\frac{20\times21\times41}{6}+20=2870+20=2890$.
Correct Answer: 2890