Let $a$ and $b$ be distinct positive real numbers. Prove by contradiction that \(\frac

Question

Let $a$ and $b$ be distinct positive real numbers. Prove by contradiction that $\frac{1}{a}+\frac{1}{b}>\frac{4}{a+b}$.[5] Give a counterexample to show that this inequality need not hold when $a$ and $b$ are nonzero but not both positive.[2]

Accepted Answer

$\frac1a+\frac1b>\frac4{a+b}$ for unequal positive $a,b$; $a=-1,\ b=-2$ gives $-\frac32<-\frac43$.

Reciprocal sum bounded by the arithmetic mean

Reciprocal sum bounded by the arithmetic mean

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