Prove $(-a+b+c)(a-b+c)(a+b-c) \leq abc$, where $a, b$ and $c$ are positive real numbers

I have tried the arithmetic-geometric inequality on $(-a+b+c)(a-b+c)(a+b-c)$ which gives

$$(-a+b+c)(a-b+c)(a+b-c) \leq \left(\frac{a+b+c}{3}\right)^3$$

and on $abc$ which gives

$$abc \leq \left(\frac{a+b+c}{3}\right)^3.$$

Since both inequalities have the same righthand side, I have tried to deduce something about the lefthand sides, but to no avail. Can somebody help me, please? I am sure it is something simple I have missed.

Solutions Collecting From Web of "Prove $(-a+b+c)(a-b+c)(a+b-c) \leq abc$, where $a, b$ and $c$ are positive real numbers"

Case 1. If $a,b,c$ are lengths of triangle.

Since
$$
2\sqrt{xy}\leq x+y\qquad
2\sqrt{yz}\leq y+z\qquad
2\sqrt{zx}\leq z+x
$$
for $x,y,z\geq 0$, then multiplying this inequalities we get
$$
8xyz\leq(x+y)(y+z)(z+x)
$$
Now substitute
$$
x=\frac{a+b-c}{2}\qquad
y=\frac{a-b+c}{2}\qquad
z=\frac{-a+b+c}{2}\qquad
$$
Since $a,b,c$ are lengths of triangle, then $x,y,z\geq 0$ and our substitution is valid. Then we will obtain
$$
(-a+b+c)(a-b+c)(a+b-c)\leq abc\tag{1}
$$

Case 2. If $a,b,c$ are not lengths of triangle.

Then at least one factor in left hand side of inequality $(1)$ is negative. In fact the only one factor is negative. Indeed, without loss of generality assume that $a+b-c<0$ and $a-b+c<0$, then $a=0.5((a+b-c)+(a-b+c))<0$. Contradiction, hence the only one factor is negative. As the consequence left hand side of inequality $(1)$ is negative and right hand side is positive, so $(1)$ obviously holds.

Here’s a geometric proof:

If $a,b,c$ satisfy the triangle inequality, let $A$ be the area of the triangle $T$ with side lengths $a$, $b$, $c$. Then the inequality reduces to
$$\frac{16A^2}{a+b+c} \leq abc$$
by Heron’s formula. Since $A$ is positive, this is equivalent to
$$\frac{2A}{a+b+c} \leq \frac{abc}{8A} \, .$$

But the left-hand side of this last inequality is the inradius of $T$ while the right-hand side is the radius of $T$’s nine-point circle. Hence the inequality follows from Feuerbach’s theorem (EDIT: or from the much simpler and more elementary argument given here).

If $a,b,c$ do not satisfy the triangle inequality, then the inequality is trivial as previously noted.

Here I give a detailed proof. Though steps could have been jumped to keep it short.

Without loss of generality we can assume that $a\ge b \ge c$

Let $$(-a+b+c)(a-b+c)(a+b-c)=S$$
$$\Rightarrow S=(-a+b+c)\{a-(b-c)\}\{a+(b-c)\}$$ $$\Rightarrow S=(-a+b+c)\{a^2-(b-c)^2\} $$ $$\Rightarrow S= (-a+b+c)\{a^2-b^2-c^2+2bc\}$$ $$\Rightarrow S=-(a^3+b^3+c^3)-2abc+b^2c+bc^2+ab^2+a^2b+ac^2+a^2c $$ $$\Rightarrow abc-S=(a^3+b^3+c^3)+3abc-(b^2c+bc^2+ab^2+a^2b+ac^2+a^2c) $$ $$\Rightarrow abc-S=(a^3-a^2b)+(b^3-b^2c)+(c^3-c^2a)+(abc-bc^2)+(abc-ab^2)+(abc-a^2c) $$$$\Rightarrow abc-S=a^2(a-b)+b^2(b-c)+c^2(c-a)+bc(a-c)+ab(c-b)+ac(b-a) $$ $$\Rightarrow abc-S=a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c-b)$$ $$\Rightarrow abc-S=(a-b)\{a(a-c)-b(b-c)\}+c(c-a)(c-b)$$ $$\Rightarrow abc-S=(a-b)^2\{a^2-b^2+c(b-a)\}+c(c-a)(c-b)$$ $$\Rightarrow abc-S=(a-b)^2\{a+b-c\}+c(c-a)(c-b)$$ Now $(c-a) \le 0$ and $(c-b) \le 0$ $$\Rightarrow c(c-a)(c-b)\ge 0 $$ and $$ (a-b)^2(a+b-c) \ge 0$$ This shows $$ abc-S \ge 0$$ $$\Rightarrow abc\ge S$$ $$\Rightarrow abc\ge (-a+b+c)(a-b+c)(a+b-c) $$

Note that no two of $(-a+b+c)$, $(a-b+c)$, and $(a+b-c)$ can be negative. If so, then one of
$$
\begin{align}
(a-b+c)+(a+b-c)&=2a\\
(a+b-c)+(-a+b+c)&=2b\\
(-a+b+c)+(a-b+c)&=2c
\end{align}
$$
would be negative, but each of $a$, $b$, and $c$ is positive. Thus, at most one can be negative. If only one were negative, then the product on the left would be non-positive and the inequality would be trivial. Therefore, we can assume that $a$, $b$, and $c$ are sides of a triangle.

By Heron’s Formula, a triangle with sides of length $a$, $b$, and $c$, has area $A$ where
$$
(a+b+c)(-a+b+c)(a-b+c)(a+b-c)=16A^2\tag{1}
$$
Let $r$ be the radius of the inscribed circle and $R$ be the radius of the circumscribed circle. Then
$$
2A=r(a+b+c)\tag{2}
$$
and
$$
4AR=abc\tag{3}
$$
Putting together $(1)$-$(3)$ yields
$$
(-a+b+c)(a-b+c)(a+b-c)R=2rabc\tag{4}
$$
The distance $d$ between the incenter and circumcenter is given by
$$
d^2=R(R-2r)\tag{5}
$$
Combining $(4)$ and $(5)$ gives
$$
\begin{align}
(-a+b+c)(a-b+c)(a+b-c)
&=\left(1-\frac{d^2}{R^2}\right)abc\\[6pt]
&\le abc\tag{6}
\end{align}
$$


Justifications of $(2)$, $(3)$, and $(5)$ can be found in this answer.

The inequality is equivalent to

$\displaystyle \sum_{\text{cyc}} a^3 + 3abc \ge \displaystyle \sum_{\text{sym}} a^2b$,

which follows directly from Schur’s Inequality.

Norbert’s answer explains the case when $a,b,c$ are not the sides of a triangle.

Let $x,y,z$ be as in this picture, where $AB=c, BC=a, CA=b$.

Then $(a,b,c)=(y+z,x+z,x+y)$.$\,$ Inequality becomes $$8xyz\le (x+y)(y+z)(z+x),$$

which as Norbert says is true by $2\sqrt{xy}\le x+y$ (proof: $\Leftrightarrow (\sqrt{x}-\sqrt{y})^2\ge 0$) and $2\sqrt{yz}\le y+z$ and $2\sqrt{zx}\le z+x$.

Some overkill approaches: by Hölder’s inequality: $$(x+y)(y+z)(z+x)\ge (\sqrt[3]{xyz}+\sqrt[3]{yzx})^3=8xyz$$

By AM-GM: $$(x+y)(y+z)(z+x)-xyz=(x+y+z)(xy+yz+zx)$$ $$\ge (3\sqrt[3]{xyz})(3\sqrt[3]{xy\cdot yz\cdot zx})=9xyz$$

By Cauchy-Schwarz: $$(z+x+y)(xy+yz+zx)\ge (\sqrt{z}\cdot \sqrt{xy} + \sqrt{x}\cdot \sqrt{yz}+\sqrt{y}\cdot \sqrt{zx})^2=9xyz$$

By Muirhead’s inequality:

$$(x+y)(y+z)(z+x)-2xyz=\sum_{\text{sym}}x^2y^1z^0\ge \sum_{\text{sym}}x^1y^1z^1=6xyz,$$

because $(2,1,0)\succ (1,1,1)$. Last one is also true by rearrangement inequality because hint: we can let WLOG $x\ge y\ge z$ and $xy \ge zx\ge yz$.