Let a,b and c be positive real numbers with $abc = 1$, prove that $\displaystyle \frac{a}{2+bc}+\frac{b}{2+ca}+\frac{c}{2+ab}\ge 1$

Let a,b and c be positive real numbers with $abc = 1$, prove that

$\displaystyle \frac{a}{2+bc}+\frac{b}{2+ca}+\frac{c}{2+ab}\ge  1$

In the problem 4 as shown in quiz 22, I asked if you could approach the problem using the Hölder's inequality, I hope you have tried it before checking out with my solution to see why the Hölder's inequality wouldn't help:

The following inequality is always true (by the Hölder's inequality) for all positive a, b and c such that $abc = 1$:

$\displaystyle \left(\sum_{\text{cyclic}}\left(\frac{(a^{\frac{1}{3}})^3}{((2+bc)^{\frac{1}{3}})^3}\right)\right)^{\frac{1}{3}}\left(\sum_{\text{cyclic}}\left((2+bc)^{\frac{1}{3}})^3\right)\right)^{\frac{1}{3}}\left(\sum_{\text{cyclic}}\left(\frac{1}{((a)^{\frac{1}{3}})^3}\right)\right)^{\frac{1}{3}}$

$\displaystyle \ge \sum_{\text{cyclic}}\left(\frac{a^{\frac{1}{3}}}{(2+bc)^{\frac{1}{3}}}\cdot (2+bc)^{\frac{1}{3}} \cdot \frac{1}{(a)^{\frac{1}{3}}}\right)$

$\displaystyle = \sum_{\text{cyclic}}\left(\frac{\cancel{a^{\frac{1}{3}}}}{\cancel{(2+bc)^{\frac{1}{3}}}}\cdot \cancel{(2+bc)^{\frac{1}{3}}} \cdot \frac{1}{\cancel{(a)^{\frac{1}{3}}}}\right)$

$\displaystyle = \sum_{\text{cyclic}}\left(1\right)$

$\displaystyle = 3$

Therefore by cubing both sides of the inequality we see that we have:

$\displaystyle \left(\sum_{\text{cyclic}}\left(\frac{(a^{\frac{1}{3}})^3}{(2+bc)^{\frac{1}{3}})^3}\right)\right)\left(\sum_{\text{cyclic}}\left((2+bc)^{\frac{1}{3}})^3\right)\right)\left(\sum_{\text{cyclic}}\left(\frac{1}{((a)^{\frac{1}{3}})^3}\right)\right)\ge 3^3$

$\displaystyle \left(\sum_{\text{cyclic}}\left(\frac{a}{2+bc}\right)\right)\left(\sum_{\text{cyclic}}\left(2+bc\right)\right)\left(\sum_{\text{cyclic}}\left(\frac{1}{a}\right)\right)\ge 3^3$

which, upon simplification, get down to

$\displaystyle \color{yellow}\bbox[5px,purple]{\left(\sum_{\text{cyclic}}\left(\frac{a}{2+bc}\right)\right)\left(6+ab+bc+ca\right)\left(ab+bc+ca\right)\ge 3^3}$

which also means, if we can prove

$\displaystyle \color{yellow}\bbox[5px,green]{\left(\sum_{\text{cyclic}}\left(\frac{a}{2+bc}\right)\right) \ge \frac{3^3}{\left(6+ab+bc+ca\right)\left(ab+bc+ca\right)}}=1$

then we're done.

But note that we haven't enough data to look for the maximum for $\displaystyle ab+bc+ca$, this suggests this method has no exit, no way out.