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What is the proof that: $$\forall \ a,b,c\geq 0:\quad a\ b\ c\geq(a-b+c)(a+b-c)(b+c-a)$$

I tried :

we can write that expression $(a-b+c)(a+b-c)(b+c-a)$ as $$-a^3+a^2 b+a^2 c+a b^2-2 a b c+a c^2-b^3+b^2 c+b c^2-c^3$$

then $$a\ b\ c\geq(a-b+c)(a+b-c)(b+c-a)$$ $$\Longleftrightarrow $$ $$a b c \geq -a^3+a^2 b+a^2 c+a b^2-2 a b c+a c^2-b^3+b^2 c+b c^2-c^3$$

Now consider $f(a)=abc-(-a^3+a^2 b+a^2 c+a b^2-2 a b c+a c^2-b^3+b^2 c+b c^2-c^3)$ for $b,c$ fixed

if i follow the metode of Quang Hoang

let $a+b-c=2x,b+c-a=2y,c+a-b=2z$, then the inequality becomes $$(y+z)(z+x)(x+y)\ge 8xyz.\tag{1}$$

for $x,y,z \geq 0$ we use AM-GM inequality $$(y+z)(z+x)(x+y)\ge 2(yz)^{1/2}(2zx)^{1/2}(2xy)^{1/2}.\tag{1}$$ but other case of $x,y,z$ is not clear.

  • i will be appreciated if someone could explain it with more detail
Educ
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2 Answers2

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Write $a+b-c=2x,b+c-a=2y,c+a-b=2z$, then the inequality becomes $$(y+z)(z+x)(x+y)\ge 8xyz.\tag{1}$$ Note that at most one of $x,y,z$ can be negative (consider the largest of $a,b,c$).

Edit: Without loss of generality, assume that $a=\max(a,b,c)$, then $$2x=(a-c)+b\ge b\ge 0, 2z=(a-b)+c\ge c\ge 0.$$

So there are two cases

  1. If $y\le0$: (1) is clear since the LHS is $\ge 0$, the RHS is $\le 0$.
  2. If $y\ge 0$, then all $x,y,z$ are $\ge 0$. Using AM-GM: $$x+y\ge 2\sqrt{xy},y+z\ge 2\sqrt{yz},z+x\ge 2\sqrt{zx}.$$ Multiplying those gives (1).
Quang Hoang
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  • From " Note that .. " until the end is less clear for me please could you expand it – Educ Nov 16 '14 at 10:02
  • you want to say that we've three cases: $(x\leq 0; y,z\geq 0 )$ or $(y\leq 0; x,z\geq 0 )$ or $(z\leq 0; x,y\geq 0 )$ let resume that cases in one case: $Max(x,y,z)<0$ – Educ Nov 16 '14 at 10:59
  • @Educ: If $a=\max(a,b,c)$, then $2x\ge b\ge 0$ and $2z\ge c\ge 0$. – Quang Hoang Nov 16 '14 at 13:04
  • i will be appreciated if you could just give more detail to your answer Thank you – Educ Nov 16 '14 at 13:24
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Let $a\geq b\geq c$.

Hence, $$abc-(a+b-c)(a+c-b)(b+c-a)=\sum_{cyc}(a^3-a^2b-a^2c+abc)=$$ $$=\sum_{cyc}a(a-b)(a-c)\geq a(a-b)(a-c)+b(b-a)(b-c)=$$ $$=(a-b)(a(a-c)-b(b-c))\geq0$$ because $a\geq b\geq0$ and $a-c\geq b-c\geq0$.

Done!

Michael Rozenberg
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