Question

Let $x_1,x_2,x_3$ be three positive numbers such that $x_1+x_2+x_3=z$.
Which of the following is/are correct?

• A. $(z-x_1)(z-x_2)(z-x_3)\le \frac{8}{27}{z}^3$
• B. $(z-x_1)(z-x_2)(z-x_3)\ge 8x_1{x}_2{x}_3$
• C. $\frac{x_2{x}_3}{x_1}+\frac{x_3{x}_1}{x_2}+\frac{x_1{x}_2}{x_3}\le z$
• D. $z^3<27x_1{x}_2{x}_3$

Answer 1

Answer: (A), (B)

The correct options are
A $(z-x_1)(z-x_2)(z-x_3)\le \frac{8}{27}{z}^3$
B $(z-x_1)(z-x_2)(z-x_3)\ge 8x_1{x}_2{x}_3$

Using $A.M.\ge G.M.$
$\frac{x_2+x_3}{2}\ge (x_2{x}_3{)}^{1/2}$
$\Rightarrow x_2+x_3\ge 2(x_2{x}_3{)}^{1/2}$

So, $(x_1+x_2)(x_2+x_3)(x_1+x_3)\ge 8x_1{x}_2{x}_3$
$\Rightarrow (z-x_1)(z-x_2)(z-x_3)\ge 8x_1{x}_2{x}_3$

Also, $\frac{(z-x_1)+(z-x_2)+(z-x_3)}{3}$
$\ge {\left[\right(z-x_1\left)\right(z-x_2\left)\right(z-x_3\left)\right]}^{1/3}\Rightarrow \frac{2z}{3}\ge \left[\right(z-x_1\left)\right(z-x_2\left)\right(z-x_3){]}^{1/3}\Rightarrow (z-x_1\left)\right(z-x_2\left)\right(z-x_3)\le \frac{8}{27}{z}^3$

$\frac{1}{2}(\frac{x_2{x}_3}{x_1}+\frac{x_1{x}_3}{x_2})\ge \sqrt{(x_3{)}^2}\Rightarrow \frac{1}{2}(\frac{x_2{x}_3}{x_1}+\frac{x_1{x}_3}{x_2})\ge x_3$
Similarly, $\frac{1}{2}(\frac{x_2{x}_3}{x_1}+\frac{x_2{x}_1}{x_3})\ge x_2$
and $\frac{1}{2}(\frac{x_1{x}_3}{x_2}+\frac{x_1{x}_2}{x_3})\ge x_1$

$\therefore \frac{x_2{x}_3}{x_1}+\frac{x_1{x}_3}{x_2}+\frac{x_1{x}_2}{x_3}\ge x_1+x_2+x_3=z$

$\frac{x_1+x_2+x_3}{3}\ge (x_1{x}_2{x}_3{)}^{1/3}\Rightarrow z^3\ge 27x_1{x}_2{x}_3$