$\frac{16}{27} \left( \frac a{b+c} + \frac b{a+c} +\frac c{a+b} \right) ^3 + \left( \frac{abc}{(a+b)(b+c)(a+c)}\right)^{\frac{1}{3}} \geq \frac52$

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.

Solution 1:

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.