Prove for all positive a,b,c $$\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b} \geq6$$
My Try
I tried taking common denominator of the expression,
$\frac{a^2b+ab^2+b^2c+c^2b+ac^2+a^2c}{abc}$
How to proceed? Is there a way to write them as perfect squares to get the least value? a Hint is much appreciated. Thanks!
On
Using Cauchy Schwarz, we can write $\displaystyle (a + b + c) \left(\frac{1}{a} +\frac{1}{b} + \frac{1}{c} \right) \geq 9$. Thus, we have, $$ \frac{a+b+c}{a} +\frac{a+b+c}{b} + \frac{a+b+c}{c} \geq 9 \implies 1+ \frac{b+c}{a} + 1 + \frac{a+c}{b} + 1 + \frac{a+b}{c} \geq 9$$ Hence, $$ \frac{a+b}{c} + \frac{b+c}{a} + \frac{c+a}{b} \geq 6 $$
On
Using Am-Gm for 6 terms we get: $$\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}\geq 6\sqrt[6]{\frac{a}{b} \frac{b}{a} \frac{b}{c} \frac{c}{b} \frac{c}{a}\frac{a}{c}}= 6$$
On
Notice that $$\frac{a}{b}+ \frac{b}{c}+ \frac{c}{a} \ge 3 \left( \frac{a}{b} \frac{b}{c} \frac{c}{a}\right)^{1/3}=3~~~~ \mbox{by AM-GM}$$ and $$\frac{b}{a}+ \frac{c}{b}+ \frac{a}{c} \ge 3 \left( \frac{b}{a} \frac{c}{b} \frac{a}{c}\right)^{1/3}=3~~~~ \mbox{by AM-GM}.$$ Adding these two we get the required inequality. In each case and overall equality occurs when $a=b=c$.
On
If we write $s=a+b+c$ then we have to prove $${s-a\over a}+{s-b\over b}+{s-c\over c} \geq 6$$ or $${s\over a}+{s\over b}+{s\over c} \geq 9$$ or $$s({1\over a}+{1\over b}+{1\over c}) \geq 9$$ or $$(a+b+c)({1\over a}+{1\over b}+{1\over c}) \geq 9$$
or $${a+b+c\over 3}\geq {3\over {1\over a}+{1\over b}+{1\over c}}$$ which is true by Am-Hm for 3 terms.
It is $$\frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{c}{a}+\frac{a}{c}\geq 6$$ and now use that $$x+\frac{1}{x}\geq 2$$ for all $$x>0$$ You can also use that $$\frac{a^2 b+ab^2+b^2c+c^2b+a^2c+ac^2}{6}\geq \sqrt[6]{(abc)^6}=abc$$