Algebraic inequalities with equality condition

Question

Let $a,b,c$ be positive real numbers such that $abc(a+b+c) =3$. Prove that $$(a+b)(b+c)(c+a)\ge 8$$ and determine when equality holds.

Answer 1

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!