This is equivalent to $(a+b+c)^3(-a+b+c)(a-b+c)(a+b-c) \leq 27a^2b^2c^2$ Let $x = -a + b + c, y = a-b+c, z = a+b-c$ and note that at most one of $x,y,z$ can be negative (since the sum of any two is positive). If $x < 0$ then $(a+b+c)^3xyz \leq 0 < 27a^2b^2c^2$. If $x,y,z > 0$, note $x+y+z = a+b+c, x+y = 2c$ etc so our inequality becomes $64xyz(x+y+z)^3 \leq 27(x+y)^2(y+z)^2(z+x)^2$ Note that $9(x+y)(y+z)(z+x) \geq 8(x+y+z)(xy+yz+zx)$ and $(xy+yz+zx)^2 \geq 3xyz(x+y+z)$. Combining these completes our proof!

I think this can also be proved using triangle. As noted in the last post, we may assume $a,b,c$ are the sides of a triangle. Multiplying LHS by $a+b+c$ and RHS by $3$, the inequality becomes $16\Delta^2 \le 3a^2b^2c^2$, where $\Delta$ is the area of the triangle. That is equivalent to $R^2\ge \frac{1}{3}$ since $\Delta = \frac{abc}{4R}$, where $R$ is the circumradius. But this is true since $R=\frac{a+b+c}{2(\sin{A}+\sin{B}+\sin{C})}$ and $\sin{A}+\sin{B}+\sin{C} \le \frac{3\sqrt{3}}{2}$ by Jensen.

Let x=3-2a y=3-2b ve z=3-2c than inequality is xyz is less than or equal to (x+z)^2(y+z)^2(x+y)^2 / 8 But we now from AM-GM xyz is less than or equal to (x+z)(y+z)(x+y) / 8 and (x+z)(y+z)(x+y) is less than or equal to (x+y+z)^3 / 3^3 . And we now that x+y+z=a+b+c=3, so that (x+z)(y+z)(x+y) is less than or equal to 1. Conclusion follows.... Sorry for my poor english........

Valentin Vornicu wrote: Let $ a,b,c$ be three positive real numbers with $ a + b + c = 3$. Prove that \[ (3 - 2a)(3 - 2b)(3 - 2c) \leq a^2b^2c^2 . \] Robert Szasz $ F(a,b,c)=a^{2}b^{2}c^{2}-(3-a)(3-b)(3-c)$ $ F(\frac{a+b}{2},\frac{a+b}{2},c)=\frac{(a+b)^{4}}{16}c^{2}-(3-(a+b))^{2}(3-2c)$ $ F(\frac{a+b}{2},\frac{a+b}{2},c)-F(a,b,c)=(a-b)^{2}[\frac{c^{2}}{16}(a^{2}+b^{2}+6ab)+2c-3]$ We have $ \frac{c^{2}}{16}(a^{2}+b^{2}+6ab)+2c-3\leq 2\frac{c^{2}}{16}(3-c)^{2}+2c-3$ we suppose c=min{a,b,c} $ \Rightarrow F(a,b,c)\geq F(x,x,c)$ (x=a+b/2) we need prove : $ F(c)=\frac{1}{16}c^{2}(3-c)^{4}-c^{2}(3-2c)\geq 0$ $ G(c)=\frac{1}{16}(3-c)^{4}-(3-2c)$ G'(c)<0 $ \Rightarrow G(c)\geq G(1)=0$ Done !

Use sin<A+sin<B+sin<C≤3√/2 in triangle. It can be the easy inequality if se put sin<B+A=sin<C... This problem is the direct consequence of that statement.

ThAzN1 wrote: This is equivalent to $$(a+b+c)^3(-a+b+c)(a-b+c)(a+b-c) \leq 27a^2b^2c^2$$

! $$(a+b)(b+c)(c+a)(b+c-a)(c+a-b)(a+b-c) \leq 8a^2b^2c^2$$https://artofproblemsolving.com/community/c6h1785721p11793688

Let $x = a+b-c$,$y = b+c-a$ and $z = c+a-b$ Note that we need to prove $xyz \le ((\frac{x+z}{2})(\frac{y+x}{2})(\frac{z+y}{2}))^2$ Note that if one of $x,y,z$ is negative then inequality is obvious so assume $x,y,z \ge 0$ and also Note that $x+y+z = 3$. Note that we need to prove $64xyz \le ((x+z)+(y+x)+(z+y))^2$ or $81xyz \le ((x+y+z)(xy+yz+zx))^2$ or $9xyz \le (xy+yz+zx)^2$ which is true since $(xy+yz+zx)^2 \ge 3(x+y+z)xyz$ and $x+y+z = 3$.