Let x and y be two positive real numbers such | vedclass

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(D) Given that $x$ and $y$ are positive real numbers such that $x+y=1$. We want to find the minimum value of $f(x, y) = \frac{1}{x} + \frac{1}{y}$. Us

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$4$

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(D) Given that $x$ and $y$ are positive real numbers such that $x+y=1$. We want to find the minimum value of $f(x, y) = \frac{1}{x} + \frac{1}{y}$. Using the Arithmetic Mean-Harmonic Mean $(AM \geq HM)$ inequality for two positive numbers $x$ and $y$: $\frac{x+y}{2} \geq \frac{2}{\frac{1}{x} + \frac{1}{y}}$ Substitute $x+y=1$ into the inequality: $\frac{1}{2} \geq \frac{2}{\frac{1}{x} + \frac{1}{y}}$ Multiplying both sides by $2(\frac{1}{x} + \frac{1}{y})$,we get: $\frac{1}{x} + \frac{1}{y} \geq 4$. Thus,the minimum value is $4$.

Let $x$ and $y$ be two positive real numbers such that $x+y=1$. Then,the minimum value of $\frac{1}{x}+\frac{1}{y}$ is

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