For the first and second we use $ \displaystyle \frac{1}{a}+\frac{1}{b}\geq\frac{4}{a+b} $ For the first and third$ \sum\frac{1}{a}\geq\frac{(1+1+1)^2}{a+b+c}=\frac{9}{a+b+c} $ And for the second and third the same$ \sum\frac{2}{a+b}\geq\frac{(3\sqrt{2})^2}{2(a+b+c)}=\frac{9}{a+b+c} $

My proof: Let $\frac{1}{a}=x$, $\frac{1}{b}=y$ and $\frac{1}{c}=z$. Hence the inequality becomes: \[x+y+z \geq \frac{2xy}{x+y}+\frac{2yz}{y+z} +\frac{2xz}{x+z}\] Since $x+y \geq \sqrt{xy}$, so: \[ \sum \frac{2xy}{x+y} \leq \sum \sqrt{xy}\] Now it becomes: $x+y+z \geq \sqrt{xy}+\sqrt{yz}+\sqrt{zx}$, which is obviously true. The next sign is a simple application of AM-HM. ( That's why i used this title.)

to prove the 1st and 3rd, i used AM-HM: $\displaystyle\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge \frac{9abc}{ab+ac+bc} \ge \frac{9}{a+b+c}$ this is because $\displaystyle\frac{abc}{ab+bc+ac} \ge \frac{1}{a+b+c+}$ since: $(abc)(a+b+c) \ge ab+bc+ac \Rightarrow a^2bc + ab^2c +abc^2 \ge ab+ac+bc \Rightarrow abc^2 \ge ab$ and goes the same for the rest of them

Your last inequality, that $abc^2\geq{ab}$ is only true if c is greater than 1. However, comparing the first and the third is immediate from AM-HM. You seemed to have forgotten to take the reciprocal. You get $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq{\frac{9}{\frac{1}{\frac{1}{a}}+\frac{1}{\frac{1}{b}}+\frac{1}{\frac{1}{c}}}}$

oh yea..... stupid me

lemma 1/a+1/b>=4/(a+b) <=> (a+b)^2>=4ab 1/a+1/b>=4/(a+b) 1/c+1/b>=4/(c+b) 1/a+1/c>=4/(a+c) <=>1/a+1/b+1/c>=2/(a+b)+2/(c+b)+2/(a+c) 2/(a+b)+2/(c+b)+2/(a+c)>=2[9/2{a+b+c}]=9/a+b+c

For the second and third, simply substitute $ x=a+b$, etc. to get $ \sum 1/x \ge 9/(x+y+z)$ which is just direct AM-HM.

Hello.

umm... Jensen sorta kills this problem without any algebraic manipulation... Just set in $ f(x)=\frac{1}{x}$, which is convex in the interval $ (0,\infty)$. $ \frac{f(a)+f(b)}{2}+\frac{f(b)+f(c)}{2}+\frac{f(c)+f(a)}{2}\geq f(\frac{a+b}{2})+f(\frac{b+c}{2})+f(\frac{c+a}{2})\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}$ For the RHS we use $ \frac{f(\frac{a+b}{2})+f(\frac{b+c}{2})+f(\frac{c+a}{2})}{3}\geq f\left(\frac{a+b+c}{3}\right)$ and the result follows.

shobber wrote: For any positive real numbers $a,b$ and $c$, prove: \[ \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \geq \dfrac{2}{a+b} + \dfrac{2}{b+c} + \dfrac{2}{c+a} \geq \dfrac{9}{a+b+c} \]

shobber wrote: For any positive real numbers $a,b$ and $c$, prove: \[ \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \geq \dfrac{2}{a+b} + \dfrac{2}{b+c} + \dfrac{2}{c+a} \geq \dfrac{9}{a+b+c} \] LOL $$\sum_{cyc}\frac{1}{2a}+\frac{1}{2b}\ge\sum\frac{2}{a+b}\ge\frac{(2+2+2)^2}{2(a+b+c)}\ge\frac{9}{a+b+c}$$