Prove that $(a+b)(b+c)(c+a) \ge8$

Question

Given that $a,b,c \in \mathbb{R}^{+}$ and $abc(a+b+c)=3$,
Prove that $(a+b)(b+c)(c+a)\ge8$.

My attempt: By AM-GM inequality, we have

$$\frac{a+b}{2}\ge\sqrt {ab} \tag{1}$$

and similarly

$$\frac{b+c}{2}\ge\sqrt {bc} \tag{2},$$ $$\frac{c+d}{2}\ge\sqrt {cd} \tag{3}.$$

Multiplying $(1)$, $(2)$ and $(3)$ together we reach

$$(a+b)(b+c)(c+a) \ge8abc.$$

Now, I need to show that $abc = 1$.

Again by AM-GM inequality we have

$$\frac{a+b+c}{3}\ge\sqrt[3]{abc} \implies \frac{\frac{3}{abc}}{3}\ge\sqrt[3]{abc} \implies abc\le1$$

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Now if we show somehow that $abc\ge1$, we are done and that's where I am stuck. Can someone please explain how to show that?

Other solutions to the above question are also welcomed.

Answer 1

If the numbers are positive here is a solution

$$3=abc(a+b+c)\ge 3abc(abc)^{\dfrac 1 3} \Rightarrow abc\le 1 \\$$

$$3=abc(a+b+c)\le \dfrac {(a+b+c)^3} {27} \cdot (a+b+c) \Rightarrow a+b+c \ge 3 \\ $$

$$ (ab+bc+ca)^2 \ge 3abc(a+b+c)=9 \Rightarrow ab+bc+ca \ge 3$$

$$(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc \ge 9-1=8$$

equality holds if and only if $a=b=c=1$

Answer 2

A geometric approach is to notice that $a,b,c>0$ and $abc(a+b+c)=3$ grant that there is a triangle with side lengths $a+b,a+c,b+c$ and area $\sqrt{3}$. Since the area of a triangle is given by the product of its side lengths divided by four times the length of the circumradius, the problem boils down to understanding what is the minimum circumradius for a given area, and it is pretty clear that the minimum is achieved by the equilateral triangle, since $$ 2R = \frac{BC}{\sin\widehat{BAC}} $$ and if a variable $A$ point travels on a line parallel to a fixed $BC$ segment (so that the area of $ABC$ is constant), the maximum $\widehat{BAC}$ angle is achieved when $BA=CA$.

Answer 3

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

Hence, we need to prove that $9uv^2-w^3\geq8$, which is a linear inequality of $v^2$

and the condition does not depend on $v^2$.

Thus it remains to prove our inequality for an extremal value of $v^2$.

$a$, $b$ and $c$ are positive roots of the equation $(x-a)(x-b)(x-c)=0$ or

$x^3-3ux^2+3v^2x-w^3=0$ or $3v^2x=-x^3+3ux^2+w^3$.

Thus, the graph of $y=3v^2x$ and the graph of $y=-x^3+3ux^2+w^3$ have three common points.

Thus, an extremal value of $v^2$ holds for the case, when a line $y=3v^2x$ is a tangent line to the graph of $y=-x^3+3ux^2+w^3$, which happens for equality case of two variables.

Id est, it remains to prove our inequality for $b=a$ and the condition gives $c=\frac{\sqrt{a^4+3}-a^2}{a}$.

Thus, we need to prove that $2a\left(a+\frac{\sqrt{a^4+3}-a^2}{a}\right)^2\geq8$, which is

$a^4-4a+3\geq0$, which is AM-GM.

Done!

Answer 4

From $abc(a+b+c)=3$ we have $$a^2+a(b+c)=\frac {3}{bc}.$$ Therefore we have $$(a+b)(b+c)(c+a)\geq 8\iff$$ $$(a+b)(a+c)\geq \frac {8}{b+c}\iff$$ $$a^2+a(b+c)\geq -bc+ \frac {8}{b+c)}\iff$$ $$(\bullet ) \quad \frac {3}{bc}\geq -bc+\frac {8}{b+c}.$$ Now $b+c\geq 2\sqrt {bc},$ so $$\frac {8}{b+c}\leq \frac {4}{\sqrt {bc}}.$$ So with $x=\sqrt {bc} $ we see that ($\bullet $ ) is satisfied if $\frac {3}{x^2}\geq -x^2+\frac {4}{x} $, or, equivalently, $\frac {3} {x^2}+x^2-\frac {4}{x}\geq 0 ,$ for all $x>0.$

We have $\frac {3}{x^2}+x^2-\frac {4}{x}=x^{-2}(x-1)(x^3+x^2+x-3). $ The terms $(x-1)$ and $(x^3+x^2+x-3)$ always have the same sign when $x\geq 0.$ (Case 1:$\;x\geq 1.$ Case 2: $\;0<x<1.$)

Inelegant compared to the answer by Booldy, but it works.