Given $ab+bc+ca =12$ then what is the least value of $a+b+c$ can we use AM - GM inequality for solving this problem
2026-03-29 02:20:47.1774750847
If $a,b,c$ are sides of a triangle then,
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$(a+b+c)^2\geq3(ab+ac+bc)=36$, which gives $a+b+c\geq6$.
The equality occurs for $a=b=c=2$.
We can prove $(a+b+c)^2\geq3(ab+ac+bc)$ by AM-GM: $$\sum_{cyc}(a^2+b^2)\geq2\sum_{cyc}ab$$ or $$\sum_{cyc}(a^2-ab)\geq0$$ or $$(a+b+c)^2\geq3(ab+ac+bc)$$