Positive integers $a, b, c$ satisfy $\frac1a +\frac1b +\frac1c<1$. Prove that $\frac1a +\frac1b +\frac1c\le \frac{41}{42}$. Also prove that equality in fact holds in the second inequality.
Problem
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Tags: inequalities, algebra
03.02.2020 09:39
How to solve this ???
03.02.2020 11:37
@above: The key restriction to this question is that $a,b,c$ are positive integers. Apply some upper bounds and you'll eventually find that "equality"; and from there, the rest will basically just be fill in.
11.02.2024 21:35
Notice that $\frac{1}{2}+\frac{1}{3}+\frac{1}{7}=\frac{41}{42}$. Without loss of generality, suppose $a\leq b\leq c$. If $a=2$ and $b=3$, then keeping in mind the first condition, $c=7$ is optimal, befitting what we want to prove. If $a=2$, $b\geq 4$ and $c \geq 5$, we get a bound weaker than the desired bound. If $a \geq 3$, $\, b \geq 3$ and $c\geq 4$, we again get a weaker bound. Hence, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\leq \frac{41}{42}$ and equality is achieved at $\left(2, 3, 7 \right)$ and its permutations.