If $ab+bc+ca=1,$ prove $1+36(abc)^2\ge\frac{21abc}{a+b+c}. $

Question

Problem. Let $a,b,c\ge 0: ab+bc+ca=1.$ Prove that:$$1+36(abc)^2\ge\frac{21abc}{a+b+c}. $$

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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.

Answer 1

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.

Answer 2

Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, we need to prove that $f(u)\geq0,$ where $$f(u)=3v^2+\frac{4w^6}{v^4}-\frac{7w^3}{u}.$$ But $f$ increases, which says that it's enough to prove our inequality for a minimal value of $u$.

Now, we know that the equation $(x-a)(x-b)(x-c)=0$ or $$x^3-3ux^2+3v^2x-w^3=0$$ or $$x^3+3v^2x-w^3=3ux^2$$ has three positive roots $a$, $b$ and $c$.

Also, let $v^2$ and $w^3$ be constants and $u$ is moved(also, $a$, $b$ and $c$ are moved).

Thus, the graph of $g(x)=x^3+3v^2x-w^3$ and the parabola $y=3ux^2$ have three common points: $(a,3ua^2),$ $(b,3ub^2)$ and $(c,3uc^2).$

Now, $g'(x)=3x^2+3v^2>0$, which says that $g$ increases, also $(0,-w^3)$ it's an inflection point of $g$ and we can draw the graph of $g$ and the parabola $y=3ux^2$.

Now, let $a<b<c$ and $u$ decreases.

Thus, $b-a$ increases and $c-b$ decreases, which says that $u$ will get a minimal value, when $c-b=0$,id est, our parabola is touched to a graph of $g$, which says that $u$ gets a minimal value for equality case of two variables.

Let $b=a$ and $c=\frac{1-a^2}{2a},$ where $0<a\leq1$.

Thus, we need to prove that: $$(3a^2-1)^2(3a^4-3a^2+1)\geq0,$$ which is obvious.

Answer 3

Proof.

I found that AM-GM also helps well.

By AM-GM,$$(b+c)^2+\left(\frac{6abc}{ab+bc+ca}\right)^2\ge \frac{12abc(b+c)}{ab+bc+ca}.$$Hence,$$a(b+c)^2+a.\left(\frac{6abc}{ab+bc+ca}\right)^2\ge \frac{12a^2bc(b+c)}{ab+bc+ca}.$$ Sum up similar inequalities, $$(a+b+c)(ab+bc+ca)+3abc+(a+b+c).\left(\frac{6abc}{ab+bc+ca}\right)^2\ge \frac{24abc(ab+bc+ca)}{ab+bc+ca}.$$The desired result follows. Equality holds at $a=b=c>0.$