If a, b, c are positive real numbers such that $a + b + c = 1$, then how do you prove that $$\frac{a} {b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{3}{2} \sqrt[3]{abc} \ge {2}.$$
Here is my proof attempt:
$$\frac{a} {b+c}+\frac{b}{c+a}+\frac{c}{a+b} = \frac{a^2} {a(b+c)}+\frac{b^2}{b(c+a)}+\frac{c^2}{c(a+b)}$$
By Cauchy-Schwarz inequality, we have
$$\frac{a^2} {a(b+c)}+\frac{b^2}{b(c+a)}+\frac{c^2}{c(a+b)} \ge \frac{(a+b+c)^2} {2(ab +bc + ca)}.$$
Since $a + b + c = 1$, this reduces to
$$\frac{a^2} {a(b+c)}+\frac{b^2}{b(c+a)}+\frac{c^2}{c(a+b)} \ge \frac{1} {2(ab +bc + ca)}.$$
And by GM – HM inequality, we have
$$\sqrt[3] {abc} \ge \frac{3abc} {ab + bc + ca}.$$
Combining these inequalities, we get
$$\frac{a} {b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{3}{2} \sqrt[3]{abc} \ge \frac{1}{2(ab+bc+ca)}+\frac{3} {2}\frac{3abc} {(ab+bc+ca)} = \frac{1+9abc} {2(ab+bc+ca)}.$$
Now, $2(ab+bc+ca)= (a+b+c)^2 – (a^2 + b^2 + c^2) = 1 – (a^2 + b^2 + c^2)$, since $a + b + c = 1$.
This implies
$$\frac{1+9abc} {2(ab+bc+ca)} = \frac{1 + 9abc} {1-(a^2 + b^2 + c^2)}.$$
We know that $$(a^2+b^2+c^2) \ge \frac{(a+b+c)^2} {3} = \frac{1} {3}.$$
Using this result, we have
$$ \frac{1 + 9abc} {1 – (a^2 + b^2 + c^2)} \ge \frac{1+9abc} {1 – \frac{1} {3}} = \frac{1+9abc} {\frac{2} {3}} = \frac{3} {2} + \frac{27} {2} abc$$
which gives us
$$\frac{a} {b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{3}{2} \sqrt[3]{abc} \ge \frac{3} {2} + \frac{27} {2} abc.$$
The $\frac{3} {2}$ term on the RHS represents the value you see in Nesbitt's inequality. The equality holds when $a = b = c = \frac{1} {3}$, but to prove that the LHS is $\ge 2$ for all values of a, b, c in the open interval (0, 1), we should have $abc \ge \frac{1} {27}$, which is clearly not true because $\max {abc} = \frac{1} {27}$, so the proof is incomplete.
Is there another way to prove this inequality?
Best Answer
Proceeding along OP's lines.
It suffices to prove that $$\frac{1+9abc} {2(ab+bc+ca)} \ge 2$$ or $$1 + 9abc - 4(ab + bc + ca) \ge 0$$ or $$(a + b + c)^3 + 9abc - 4(ab + bc + ca)(a + b + c) \ge 0$$ which is degree three Schur i.e. $a(a-b)(a-c) + b(b-c)(b-a) + c(c-a)(c-b) \ge 0$.