I don't quite remember where this problem is from. I came across is sometime last summer, when I was in an olympiad-problem mood and I decided to improve my inequality skills.

Suppose $a,b,c > 0$. Then we want to show that

$$\frac{16}{27} \left( \frac{a}{b+c} + \frac{b}{a+c} +\frac{c}{a+b} \right) ^3 + \left( \frac{abc}{(a+b)(b+c)(a+c)}\right)^{\frac{1}{3}} \geq \frac{5}{2}$$

I think that there are many things to notice. Firstly, it's homogenous. The left part is tantalizingly close to Nesbitt's inequality. The right part seems to demand AM-GM attention.

  • $\begingroup$ I think the right part has to be $\frac{(a+b)(b+c)(a+b)}{abc}$not $\frac{abc}{(a+b)(b+c)(a+b)} $ $\endgroup$ – Babak Miraftab Jan 31 '12 at 7:03
  • 2
    $\begingroup$ And I think it should be $\frac{abc}{ \textbf{(a+c)}(b+c)(a+b)}$. $\endgroup$ – Gigili Jan 31 '12 at 8:38
  • $\begingroup$ Shouldn't the factors of the denominator of the right-hand term be the denominators of the left-hand term: namely $b+c$, $c+a$, and $a+b$ ? $\endgroup$ – John Bentin Jan 31 '12 at 9:27
  • $\begingroup$ I would like to have a "certified" version of the inequality. Indeed the left term is greater or equal to 2 by Nesbitt, but the right term is lower or equal than $frac{1}{2}$ instead of being greater or equal. On the other hand I didn´t find a counterexample, so I think that the two terms "interact" for giving the inequality. On the right hand side the denominator should be as John pointed out, this is why I would like the confirmation it is the only misprint there. $\endgroup$ – Giovanni De Gaetano Jan 31 '12 at 14:40
  • $\begingroup$ Whoops! Yes, I'm sorry about the $a + b$ twice! $\endgroup$ – davidlowryduda Jan 31 '12 at 15:12

First, make the substitutions $$ x= \frac{a}{b+c}, \quad y= \frac{b}{a+c}, \quad z= \frac{c}{a+b}. $$ The strategy will be to reduce the problem to an inequality in the single variable $t=(xyz)^{1/3}$. Note that $xy+yz+xz+2xyz=1$, and the inequality to be proved is $$ \frac{16}{27}\left(x+y+z\right)^3+(xyz)^{1/3}\geq\frac{5}{2}. $$ Now $$ \frac{(x+y+z)^2}{3}\geq xy+yz+xz=1-2xyz $$ and also $x+y+z\geq3/2$ by Nesbitt's inequality. Therefore, $$ \frac{16}{27}(x+y+z)^3=\frac{16}{9}\cdot(x+y+z)\cdot\frac{(x+y+z)^2}{3}\geq\frac{8}{3}(1-2xyz), $$ and it is sufficient to prove the inequality $$ \frac{8}{3}(1-2xyz)+(xyz)^{1/3}\geq\frac{5}{2}. $$ Now $xyz\leq1/8$, because AM-GM gives $8abc\leq (a+b)(b+c)(a+c)$ by grouping pairs on the right-hand side (e.g., $2abc\leq a^2b+bc^2$). Thus by setting $t=(xyz)^{1/3}$, we are reduced to proving that the polynomial $$ f(t):= 8\left(\frac{1-2t^3}{3}\right)+t-\frac{5}{2} = \frac{1}{6}+t-\frac{16}{3}t^3. $$ is nonnegative for $t\in[0,1/2]$. Since $f(0)>0$ and $f(1/2)=0$, we can show that $f(c)>0$ whenever $c$ is a critical point of $f$. But $f'(t)=1-16t^2$, which has $c=1/4$ as its only zero in $[0,1/2]$. As $f(1/4)=4/12>0$, we are done.

  • $\begingroup$ Nice! That construction is very clean (or as clean as these things get!) $\endgroup$ – Elchanan Solomon Feb 1 '12 at 2:24

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