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.
Best Answer
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.