# 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.

• In general, if $x < z$ and $y < z$ there is no reason to believe that $x < y$ or $y < x$. – TMM Jul 14 '12 at 17:57
• think of $a, b$ and $c$ as lengths of a triangle. – Mohamez Jul 14 '12 at 18:10
• @mohamez: below, I derive a simple formula for the ratio of the two sides of this inequality which involves only the circumradius and the distance between the incenter and circumcenter. – robjohn Jul 15 '12 at 17:38
• You cannot apply AM-GM on $-a+b+c, a-b+c, a+b-c$, unless you assume or prove that $a,b,c$ are the sides of a triangle (so that the terms are all nonnegative). – user26486 Jun 3 '15 at 18:29

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.

• That does it for all. If one of the three terms on the left is negative, the inequality is trivial. And we cannot have two ngative, say the first two, for then their sum $2c$ would be negative. – André Nicolas Jul 14 '12 at 18:46
• $a,b,c$ are lengths of a triangle if and only if $(-a+b+c),(a-b+c),(a+b-c)\ge 0$. Otherwise, without loss of generality, let $a>b+c$. Then $(a-b+c),(a+b-c)>0$. Thus $(-a+b+c)(a-b+c)(a+b-c)<0\le a b c$. – precarious Jul 14 '12 at 18:47
• And if they are not the lengths of a triangle, then exactly one of the three factors on the left-hand side is negative, so the inequality trivially holds. – Théophile Jul 14 '12 at 18:48
• While I was writting this addition you gave exactly the same answer. – Norbert Jul 14 '12 at 18:49
• (Comment now seen to be covered by other comments) – John Bentin Jul 14 '12 at 19:35

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.

• You should explain more, e.g. what your sum is over, how you derive the equivalence etc. – Julian Kuelshammer Oct 22 '12 at 6:35
• @JulianKuelshammer It is standard to denote cyclic sums by $\sum_{\text{cyc}}$ and symmetric sums by $\sum_{\text{sym}}$, so he means $a^3+b^3+c^3+3abc\ge a^2b+a^2c+b^2a+b^2c+c^2a+c^2b$. To derive the equivalence, just expand. Schur's inequality is very simple to prove (proof in Wikipedia). – user26486 Jun 3 '15 at 18:38

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

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

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 $xy \ge zx\ge yz$.

• This is the best solution on this page. – Nairit Sarkar Jul 5 '18 at 2:29

A proof using the order inequality:

As @robjon points out, no two of $$-a+b+c, a-b+c, a+b-c$$ can be negative.
If exactly one of them is negative, the inequality is trivial. Hence, assume that $$-a+b+c, a-b+c, a+b-c > 0$$.

Now, it's not difficult to see that $$(b+c, a+c, a+b)$$ and $$(-a, -b, -c)$$ are equally ordered.
And as $$-a+b+c, a-b+c, a+b-c > 0$$, it follows from the order inequality for products that $$(-a+b+c)(a-b+c)(a+b-c)\le(b+c-b)(a+c-c)(a+b-a)=abc$$