Inequality of two GMs whose AMs are same

Question

Assume that $a$, $b$ and $c$ are positive and sides of a triangle.

We know that:

$\frac{a + b + c}{3} \geq (abc)^{1/3}$ and $\frac{(a + b - c) + (b + c - a) + (c + a - b))}{3} \geq ((a + b - c)(b + c - a)(c + a - b))^{1/3}$

Here $(a + b - c) + (b + c - a) + (c + a - b) = a + b + c$

Now which one is bigger?

$abc$ or $(a + b - c)(b + c - a)(c + a - b)$

Update

As @Aqua suggested, It seems that if $a$, $b$ and $c$ are not sides of a triangle then later multiplication is non-positive.

@aqua That is not necessary! But the overall sum or multiplication will be positive. — Jayram Kumar, Aug 25 '19 at 07:15
It is not enought! But if you say that a,b,c are sides of a triangle, that would do. — nonuser, Aug 25 '19 at 07:16
@Aqua Please, I am waiting for an answer from you. I have already updated the question as per your advice. — Jayram Kumar, Aug 25 '19 at 07:43
If you just want to compare $abc$ and $(a + b - c)(b + c - a)(c + a - b)$ then you don't need to be the sides of a triangle, only if you want to use Am-Gm — nonuser, Aug 25 '19 at 07:54

score 2 · Accepted Answer · answered Aug 25 '19 at 08:05

2

Using the Ravi substitution we get $$ \left( y+z \right) \left( x+z \right) \left( x+y \right) -8\,zxy\geq 0$$ and this is AM-GM.

answered Aug 25 '19 at 08:05

Dr. Sonnhard Graubner

97,058

Nice! +1, just write what is Ravi substitution – nonuser Aug 25 '19 at 08:08
The Ravi substitution use $$a=y+z,b=x+z,c=x+y$$ – Dr. Sonnhard Graubner Aug 25 '19 at 08:55
See here for further details https://brilliant.org/wiki/ravi-substitution/ – Dr. Sonnhard Graubner Aug 25 '19 at 08:57
$x = \frac{b + c - a}{2}$, $y = \frac{c + a - b}{2}$ and $z = \frac{a + b - c}{2}$ – Jayram Kumar Aug 25 '19 at 09:43

Michael Rozenberg · Answer 2 · 2019-08-25T07:55:37.687

For a triangle we have:$$abc-\prod_{cyc}(a+b-c)=\sum_{cyc}(a^3-a^2b-a^2c+abc)=\frac{1}{2}\sum_{cyc}(2a^3-2a^2b-2a^2c+2abc)=$$ $$=\frac{1}{2}\sum_{cyc}(2a^2-2abc-(2a^2c+b^2c-4abc))=$$ $$=\frac{1}{2}\sum_{cyc}(a-b)^2(a+b+c-2c)=\frac{1}{2}\sum_{cyc}(a-b)^2(a+b-c)\geq0.$$

For any non-negative $a$, $b$ and $c$ we can continue: $$abc-\prod_{cyc}(a+b-c)=\frac{1}{2}\sum_{cyc}(a-b)^2(a+b-c)\geq$$ $$\geq\frac{1}{2}((a-c)^2(a+c-b)+(b-c)^2(b+c-a))\geq$$ $$\geq\frac{1}{2}((b-c)^2(a+c-b)+(b-c)^2(b+c-a))=(b-c)^2c\geq0$$ because we can assume before that $a\geq b\geq c$.

Inequality of two GMs whose AMs are same

2 Answers2