1
$\begingroup$

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

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

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.

$\endgroup$
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $$a=b=c$$

$\endgroup$
  • $\begingroup$ 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 $a = b = c$. $\endgroup$ – Darth Geek May 2 '17 at 18:00
  • $\begingroup$ yes this is a good idea, indeed $\endgroup$ – Dr. Sonnhard Graubner May 2 '17 at 18:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.