Proving $1/(a+b) + 1/(b+c) + 1/(c+a) > 3/(a+b+c)$ for positive $a, b, c\,$?

Solution 1:

Since $a > 0$, $b > 0$, and $c > 0$, we have $$ 0 < a +b < a+b + c, $$ and therefore upon dividing both sides of this inequality by $(a+b)(a+b+c) > 0$, we get $$ 0 < \frac{1}{a+b+c} < \frac{1}{a+b}, $$ which implies $$ \frac{1}{a+b} > \frac{1}{a+b+c}. \tag{1} $$ Similarly, we have the inequalities $$ \frac{1}{b+c} > \frac{1}{a+b+c} \tag{2} $$ and $$ \frac{1}{c+a} > \frac{1}{a+b+c}. \tag{3} $$ Using (1), (2), and (3), we obtain \begin{align} \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a } &> \frac{1}{a + b + c} + \frac{1}{a + b + c} + \frac{1}{a + b + c} \\ &= \frac{3}{a + b + c}, \end{align} as required.

Solution 2:

By AM-HM inequality $$\frac13 \left(\frac1{a+b}+\frac1{b+c}+\frac1{c+a}\right)\ge \frac{3}{(a+b)+(b+c)+(c+a)}.$$ Hence $\displaystyle \frac1{a+b}+\frac1{b+c}+\frac1{c+a}\ge \frac{9}2\frac1{a+b+c}$. Because $\dfrac92>3$ the problem is solved.

Solution 3:

Other solution is Cauchy-Schwarz, that can be applied in one of its several versions.

• This is the standard one:

$[(\frac{1}{\sqrt{a+b}})^2+(\frac{1}{\sqrt{b+c}})^2+(\frac{1}{\sqrt{c+a}})^2][((\sqrt{a+b})^2+(\sqrt{b+c})^2+(\sqrt{c+a})^2]\ge 3^2$

From here the factor $9/2$, as already found, follows.

• Titu's inequality https://brilliant.org/wiki/titus-lemma/#

Here we directly have:

$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{(1+1+1)^2}{2(a+b+c)}$

• Holders form: https://brilliant.org/wiki/holders-inequality/ :

$(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})^{1/2}((a+b)+(b+c)+(c+a))^{1/2}\ge 1^{1/2}+1^{1/2}+1^{1/2}$

I wrote this answer also for a reference of mine to the different forms of C.S.