The inequivality ${1\over a}+{1\over b}\ge{4\over a+b}$, which is equivalent with $(a-b)^2\ge0$ after simple algebra, holds, so if we sum three these after cyclic changes of variables, we (after dividing by 2) get ${1\over a} +{1\over b}+{1\over c}\ge{2\over a+b}+{2\over b+c}+{2\over c+a}$, which is exactly the first part of inequality. For the second part, we'll use the lemma, sometimes called T2's lemma, SQ lemma, or (I prefer this) CS fractionfighter. If you don't know any of this names, then just search for some slightly modified CS inequality in this : \[{2\over a+b}+{2\over b+c}+{2\over c+a}={4\over 2a+2b}+{4\over 2b+2c}+{4\over 2c+2a}\ge{(2+2+2)^2\over4a+4b+4c}={9\over a+b+c}\] which is exactly second part of inequality. Q.E.D.

Alternate for the second part: By AM, HM, $\frac{(a+b)+(b+c)+(c+a)}3 \ge \frac{3}{\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}$ $ \frac{2(a+b+c)}3 \ge \frac{3}{\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}}$ $ \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \ge \frac{9}{2(a+b+c)}$ $ \frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a} \ge \frac{9}{a+b+c}$

For left part: Let's make a substition such that \[\frac{1}{a}=x,\frac{1}{b}=y,\frac{1}{c}=z\]Then the inequality takes the form \[x+y+z\geq\frac{2xy}{x+y}+\frac{2xz}{x+z}+\frac{2yz}{y+z}\]We have \[\sum{\frac{2xy}{x+y}\le\sum{\frac{2xy}{2\sqrt{xy}}}=\sum{\sqrt{xy}}}\le\sum{\frac{x+y}{2}}=x+y+z\] It is easy to solve right part

djb86 wrote: For any positive numbers $a, b, c$ prove the inequalities \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge \frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}\ge \frac{9}{a+b+c}.\] Ireland 1998 Q7

2nd side is TITU 1st side By homogenity a+b+c=1 Then 1/a+1/b+1/c>=9 9>=2/a+b +2/b+c +2/c+a 9/2>=1/1-c +1/1-b +1/1-a F(x)=1/1-x F''(x)<0 Then 27/2>=3*f(1/3)=27/2