If $a,b,c$ are positive, prove that $\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a} \geq \frac{9}{a+b+c}$

If $a,b,c$ are positive real numbers, prove that $$\frac{2}{a+b}+\frac{2}{b+c}+ \frac{2}{c+a}≥ \frac{9}{a+b+c}$$

• AM-HM after an easy substitution – Exodd May 2 '17 at 15:33

Let $a+b+c = s$. Then we have to prove

$$\dfrac{1}{s-a} + \dfrac{1}{s-b} + \dfrac{1}{s-c} \geq \dfrac{9}{2s},$$

or, equivalently,

$$\dfrac{3}{\dfrac{1}{s-a} + \dfrac{1}{s-b} + \dfrac{1}{s-c}} \leq \dfrac{2s}{3}.$$

Note that the LHS is the harmonic mean of $s-a,s-b,s-c$ and the RHS is the arithmetic mean of the same numbers. This inequality is true by the AM-HM inequality.

By Cauchy-Schwartz. $$\sum_{cyc}\frac{1}{a+b}\sum_{cyc}(a+b)\geq (1+1+1)^2=9\implies 2\cdot\sum_{cyc}\frac{1}{a+b}\geq2\cdot \frac{9}{\sum_{cyc}(a+b)}=\frac{9}{a+b+c}$$

your inequality is equivalent to $$2\,{a}^{3}-{a}^{2}b-{a}^{2}c-a{b}^{2}-a{c}^{2}+2\,{b}^{3}-{b}^{2}c-b{c }^{2}+2\,{c}^{3}>0 >0$$ after Clearing the denominators and this is equivalent to $$(a-b)(a^2-b^2)+(a-c)(a^2-c^2)+(b-c)(b^2-c^2)\geq 0$$ which is true. the equal sign holds if $$a=b=c$$

• It might be a good idea to rewrite the LHS as $$\sum_{cyc} (a-b)^2(a+b)$$ to make it clearer that is non-negative and that indeed the equal holds iff $a = b = c$. – Darth Geek May 2 '17 at 18:00
• yes this is a good idea, indeed – Dr. Sonnhard Graubner May 2 '17 at 18:01