Let $x$ and $y$ be positive real numbers such that $x\ne y$. Prove by contradiction

Question

Let $x$ and $y$ be positive real numbers such that $x\ne y$. Prove by contradiction that $\frac{x}{y}+\frac{y}{x}>2$.[5] Give a counterexample to show that this inequality may not hold if $x$ and $y$ are not both positive.[2]

Accepted Answer

$\frac{x}{y}+\frac{y}{x}>2$ for unequal positive $x,y$; $x=1,\ y=-1$ gives $-2\leq 2$.

Reciprocal inequality for distinct positive numbers

Reciprocal inequality for distinct positive numbers

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