Let $a,b,c$ be positive real numbers. $a+b+c=\frac 1a +\frac 1b +\frac 1c$ then prove that $ab+bc+ca \geq 3$
Using CS Inequality $(a+b+c)(\frac 1a +\frac 1b +\frac 1c) >9$. then by hyp $\frac 1a +\frac 1b +\frac 1c >3$. Now can we prove that $abc>1$?
Best Answer
The condition gives $$1=\frac{abc(a+b+c)}{ab+ac+bc}$$ Id est, we need to prove that $$ab+ac+bc\geq\frac{3abc(a+b+c)}{ab+ac+bc}$$ or $$(ab+ac+bc)^2\geq3abc(a+b+c)$$ or $$\sum_{cyc}(a^2b^2+2a^2bc)\geq3\sum_{cyc}a^2bc$$ or $$\sum_{cyc}c^2(a-b)^2\geq0.$$
If you want to prove that $abc\geq1$ then it's impossible because it's wrong.
Indeed, $$abc\geq1$$ it's $$abc\geq\left(\sqrt{\frac{abc(a+b+c)}{ab+ac+bc}}\right)^3$$ or $$ab+ac+bc\geq(a+b+c)\sqrt[3]{abc},$$ which is wrong for $a\rightarrow+\infty.$