# $(a+b-c)(b+c-a)(c+a-b) \leq abc$

I think i have proved $(a+b-c)(b+c-a)(c+a-b) \leq abc$, bu would like some clarification if this is a valid proof, as I have some doubts.

Firstly, expanding the LHS we obtain,

$-a^3+a^2b+a^2c+ab^2-2abc+ac^2-b^3+b^2c+bc^2-c^3\leq abc$

Dividing both sides by $-1$

$-abc\leq a^3-a^2b-a^2c-ab^2+2abc-ac^2+b^3-b^2c-bc^2+c^3$

Hence

$0\leq a^3-a^2b-a^2c-ab^2+3abc-ac^2+b^3-b^2c-bc^2+c^3$

Now, we find:

$a^2b+a^2c+ab^2+ac^2+b^2c+bc^2\leq a^3+b^3+c^3+3abc$

Now, I know Schurs inequality states for non negative real numbers a, b c and p:

$a^p(a-b)(a-c)+b^p(b-c)(b-a)+c^p(c-b)(c-a) \geq 0$

So, setting p=1, we find that

$a^3+b^3+c^3+3abc\geq ab(a+b)+ac(a+c)+bc(b+c)$

Which simplifies down to:

$a^2b+a^2c+ab^2+ac^2+b^2c+bc^2\leq a^3+b^3+c^3+3abc$

Which is the condition I was aiming to prove, hence:

$(a+b-c)(b+c-a)(c+a-b) \leq abc$

• I visited your previous question, and not much has changed from there, so this is correct. I will try to look for an alternate proof, nevertheless. – астон вілла олоф мэллбэрг Oct 14 '17 at 9:18
• what kind of numbers are $a,b,c$? – Dr. Sonnhard Graubner Oct 14 '17 at 9:21
• Sorry, a, b, c are positive real numbers. – Sam Gregg Oct 14 '17 at 11:34

.Ok, here is a proof by Ravi's substitution, which makes things ever so simple.

Suppose that $$a \geq b \geq c > 0$$ WLOG. Two cases arise : $$a \geq b+c$$ and $$a < b+c$$.

Suppose the first is true. Then, the RHS is non-positive (two positive and one non-positive term), while the RHS is strictly positive, so it follows.

Suppose $$a. Then, $$a,b,c$$ form a triangle, so by Ravi's substitution, $$a=x+y,b=y+z,c=x+z$$ where $$x,y,z$$ are some positive real numbers.

Note that:$$(a+b-c)(a-b+c)(b+c-a) = 8xyz$$, while $$abc = (x+y)(y+z)(x+z)$$. Now, the inequality follows by AM-GM: $$x+y \geq 2\sqrt{xy}$$ etc. and multiply all these together to get the result.

if $a,b,c$ are the side length of a triangle we have $$R\geq 2r$$ by Euler, with $$A=\frac{abc}{4R}$$ and $$A=sr$$ we get $$sabc\geq 8A^2$$ with $$A=\sqrt{s(s-a)(s-b)(s-c)}$$ we get $$sabc\geq s(s-a)(s-b)(s-c)$$ with $$s=\frac{a+b+c}{2}$$ we obtain $$sabc\geq s(s-a)(s-b)(s-c)$$ or $$abc\geq (-a+b+c)(a-b+c)(a+b-c)$$

I think it's not proof because you proved the Schur's inequality by the same Schur's inequality.

The proof of your inequality for non-negative variables.

Since our inequality is symmetric, we can assume that $a\geq b\geq c$.

Thus, $$abc-(a+b-c)(a+c-b)(b+c-a)=\sum_{cyc}(a^3-a^2b-a^2c+abc)=$$ $$=\sum_{cyc}a(a-b)(a-c)\geq a(a-b)(a-c)+b(b-a)(b-c)=$$ $$=(a-b)(a(a-c)-b(b-c))\geq0.$$ Done!