Question

If $f: R \rightarrow R^+$ is differentiable function such that $|f(y) \cdot f(x-y) + (x-y)^2| \geq |f(x)f(x-y)|\forall x, y \in R$, $f(0) = 5$, then the value of $f(5)$ is

Answer 1

5

Answer 2

Let the given inequality be $|f(y) \cdot f(x-y) + (x-y)^2| \geq |f(x)f(x-y)|$ for all $x, y \in R$. We are given that $f: R \rightarrow R^+$ is a differentiable function, which means $f(z) > 0$ for all $z \in R$. Since $f(x-y) > 0$, we have $|f(x-y)| = f(x-y)$. We can divide both sides by $f(x-y)$: $|f(y) + \frac{(x-y)^2}{f(x-y)}| \geq |f(x)|$. Since $f(x) > 0$, $|f(x)| = f(x)$. $|f(y) + \frac{(x-y)^2}{f(x-y)}| \geq f(x)$. Since $f(y) > 0$ and $(x-y)^2 \geq 0$ and $f(x-y) > 0$, the term $f(y) + \frac{(x-y)^2}{f(x-y)}$ is always non-negative. So, $|f(y) + \frac{(x-y)^2}{f(x-y)}| = f(y) + \frac{(x-y)^2}{f(x-y)}$. The inequality simplifies to $f(y) + \frac{(x-y)^2}{f(x-y)} \geq f(x)$ for all $x, y \in R$.

This inequality must hold for all $x, y \in R$. Let's consider specific values for $y$. Case 1: Let $y=x$. Substituting $y=x$ into the inequality, we get $f(x) + \frac{(x-x)^2}{f(x-x)} \geq f(x)$, which simplifies to $f(x) + 0 \geq f(x)$, or $f(x) \geq f(x)$. This is always true and provides no specific information about $f(x)$.

Case 2: Let $y=0$. Substituting $y=0$ into the inequality, we get $f(0) + \frac{(x-0)^2}{f(x-0)} \geq f(x)$, which simplifies to $f(0) + \frac{x^2}{f(x)} \geq f(x)$ for all $x \in R$. We are given $f(0)=5$. So, $5 + \frac{x^2}{f(x)} \geq f(x)$ for all $x \in R$.

Let's rearrange this inequality: $5 \geq f(x) - \frac{x^2}{f(x)}$ $5 \geq \frac{[f(x)]^2 - x^2}{f(x)}$. Since $f(x) > 0$, we can multiply by $f(x)$ without changing the direction of the inequality: $5f(x) \geq [f(x)]^2 - x^2$ $[f(x)]^2 - 5f(x) - x^2 \leq 0$ for all $x \in R$.

Let's consider the original inequality $f(y) + \frac{(x-y)^2}{f(x-y)} \geq f(x)$ again. Let $x$ be fixed. Consider the function $g(y) = f(y) + \frac{(x-y)^2}{f(x-y)}$. The inequality states that $g(y) \geq f(x)$ for all $y$. This means that the minimum value of $g(y)$ with respect to $y$ is greater than or equal to $f(x)$. If the minimum of $g(y)$ occurs at some point $y_0$, then $g'(y_0)=0$. Let's calculate $g'(y)$: $g'(y) = \frac{d}{dy} \left( f(y) + \frac{(x-y)^2}{f(x-y)} \right)$. Using the chain rule for the second term, let $u = x-y$, so $\frac{du}{dy} = -1$. $\frac{d}{dy} \left( \frac{(x-y)^2}{f(x-y)} \right) = \frac{d}{du} \left( \frac{u^2}{f(u)} \right) \frac{du}{dy} = \left( \frac{2u f(u) - u^2 f'(u)}{[f(u)]^2} \right) (-1)$ $= - \frac{2(x-y) f(x-y) - (x-y)^2 f'(x-y)}{[f(x-y)]^2} = \frac{(x-y)^2 f'(x-y) - 2(x-y) f(x-y)}{[f(x-y)]^2}$. So, $g'(y) = f'(y) + \frac{(x-y)^2 f'(x-y) - 2(x-y) f(x-y)}{[f(x-y)]^2}$.

Consider the specific point $y=x$. $g(x) = f(x) + \frac{(x-x)^2}{f(x-x)} = f(x) + \frac{0}{f(0)} = f(x)$. Since $g(y) \geq f(x)$ for all $y$, and $g(x) = f(x)$, this implies that $y=x$ is a point where the function $g(y)$ attains its minimum value, which is exactly $f(x)$. For $y=x$ to be a minimum point, the derivative $g'(y)$ must be zero at $y=x$. $g'(x) = f'(x) + \frac{(x-x)^2 f'(x-x) - 2(x-x) f(x-x)}{[f(x-x)]^2} = f'(x) + \frac{0 \cdot f'(0) - 0 \cdot f(0)}{[f(0)]^2} = f'(x) + 0 = f'(x)$. Since $y=x$ is the minimum point, $g'(x)$ must be zero. $g'(x) = 0 \implies f'(x) = 0$ for all $x \in R$.

If $f'(x) = 0$ for all $x \in R$, then $f(x)$ must be a constant function. Let $f(x) = c$. Since $f(0)=5$, the constant must be $c=5$. So, $f(x)=5$.

Let's verify if $f(x)=5$ satisfies all the given conditions.

1. $f: R \rightarrow R^+$: $f(x)=5$ maps $R$ to $\{5\}$, which is a subset of $R^+$. This is satisfied.
2. $f$ is differentiable: $f(x)=5$ is a constant function, and its derivative $f'(x)=0$ exists for all $x \in R$. This is satisfied.
3. $f(0)=5$: $f(0)=5$ is given and satisfied by $f(x)=5$.
4. $|f(y) \cdot f(x-y) + (x-y)^2| \geq |f(x)f(x-y)|$ for all $x, y \in R$: Substitute $f(x)=5$ into the inequality: $|5 \cdot 5 + (x-y)^2| \geq |5 \cdot 5|$ $|25 + (x-y)^2| \geq |25|$ $|25 + (x-y)^2| \geq 25$. Since $(x-y)^2 \geq 0$, $25 + (x-y)^2 \geq 25$. The term $25 + (x-y)^2$ is always non-negative, so $|25 + (x-y)^2| = 25 + (x-y)^2$. The inequality becomes $25 + (x-y)^2 \geq 25$, which simplifies to $(x-y)^2 \geq 0$. This is true for all $x, y \in R$. So, $f(x)=5$ satisfies the given inequality for all $x, y \in R$.

Since $f(x)=5$ is the unique function satisfying the given conditions, we can find $f(5)$. $f(5) = 5$.