Problem. Let $a,b,c\ge 0: ab+bc+ca=1.$ Prove that:$$1+36(abc)^2\ge\frac{21abc}{a+b+c}. $$
I think the problem is not hard but I am still stuck to find nice proofs.
After homogenizing, we need to prove
$$ab+bc+ca+\left(\frac{6abc}{ab+bc+ca}\right)^2\ge \frac{21abc}{a+b+c}.\tag{*}$$
Here is an SOS expression (my program is not good enough)
$$ LHS-RHS= \frac{\sum_{cyc} \left[3a^2 b^2 c (a-b)^2 + c^3 (a-b)^4\right]}{(ab+bc+ca)^2 (a+b+c)}. $$
Also, I've tried pqr method.
WLOG, assume that $ab+bc+ca=3.$ Now, $(*)$ becomes notation expression$$3+4r^2-\frac{21r}{p}\ge 0 \iff 3p+4pr^2-21r\ge 0.$$Consider the function by $r$ $$f(r)=4p.r^2-21r+3p, \forall 0<r\le \dfrac{p}{3},$$ Now, we obtain: $f'(r)=8pr-21.$ My strategy splits two cases.
- $f'(r)\ge 0\Longleftrightarrow \dfrac{3}{p}\ge r\ge \dfrac{21}{8p},$ hence $f(r)$ is increasing.
By using Schur of third degree \begin{align*} f(r) & \ge f\left(\frac{4(12-p^2)}{9}\right)\\ &=4p.\left(\frac{p(12-p^2)}{9}\right)^2-21\left(\frac{p(12-p^2)}{9}\right)+3p\ge 0 \\ &\Longleftrightarrow \frac{4}{81}p^7-\frac{32}{27}p^5+\frac{85}{9}p^3-25p\ge 0\\ &\iff p(p+3)(p-3)(15-2p^2)^2\ge 0, \end{align*} which is obvious by $p\ge 3.$
I stop here with the remain case $0\le r< \dfrac{21}{8p}.$
Hope you can help me continue my idea. Also, all solution and idea is welcome.
Best Answer
pqr method:
Let $p = a + b + c, q = ab + bc + ca = 1, r =abc$.
We need to prove that $$1 + 36r^2 \ge \frac{21r}{p}$$ which is written as $$36\left(r - \frac{7}{24p}\right)^2 + \frac{16p^2 - 49}{16p^2} \ge 0.\tag{1}$$
If $p \ge 7/4$, we have $\frac{16p^2 - 49}{16p^2}\ge 0$. (1) is true.
If $p < 7/4$, using $r \ge \frac{4pq - p^3}{9}$ (degree three Schur) and $p^2 \ge 3q = 3$, we have $$r - \frac{7}{24p} \ge \frac{4pq - p^3}{9} - \frac{7}{24p} = -\frac{8p^4 - 32p^2 + 21}{72p} > 0.$$
Thus, it suffices to prove that $$36\left(\frac{4pq - p^3}{9} - \frac{7}{24p}\right)^2 + \frac{16p^2 - 49}{16p^2} \ge 0$$ or $$\frac19(p^2 - 3)(2p^2 - 5)^2 \ge 0$$ which is true using $p^2 \ge 3q = 3$.
We are done.