Question

Let $x,y,z$ be positive reals. Which of the following implies $x=y=z?$
(I) $x^3+y^3+z^3=3xyz$
(II) $x^3+y^{2}z+y{z}^2=3xyz$
(II) $x^3+y^{2}z+z^{2}x=3xyz$
(II) $(x+y+z{)}^3=27xyz$

• A. I, IV only
• B. I, II, IV only
• C. I, II and III only
• D. All of them

Answer 1

The correct option is B I, II, IV only

(I) $x^3+y^3+z^3=3xyz$
$\Rightarrow x^3+y^3+z^3-3xyz=0$
$\Rightarrow (x+y+z)(x^2+y^2+z^2-xy-yz-zx)=0$
$\Rightarrow \frac{1}{2}(x+y+z)(2x^2+2y^2+2z^2-2xy-2yz-2zx)=0$
$\Rightarrow \frac{1}{2}(x+y+z)\left[\right(x-y{)}^2+(y-z{)}^2+(z-x{)}^2]=0$
Since $x+y+z\ne 0,$
$(x-y{)}^2+(y-z{)}^2+(z-x{)}^2=0$
$\Rightarrow x=y=z$

Alternate: AM $\ge$ GM
$\Rightarrow \frac{x^3+y^3+z^3}{3}\ge \sqrt[3]{3x^3{y}^3{z}^3}$
$\Rightarrow x^3+y^3+z^3\ge 3xyz$
$\text{But, if}{x}^3+y^3+z^3=3xyz$
$\Rightarrow x=y=z[\because \text{AM=GM}]$

(II) $\frac{x^3+y^{2}z+y{z}^2}{3}\ge \sqrt[3]{3x^3{y}^{2}zy{z}^2}$
$\Rightarrow x^3+y^{2}z+y{z}^2\ge 3xyz$
$\text{Then,}{x}^3+y^{2}z+y{z}^2=3xyz\Rightarrow x=y=z$

(III) $\text{Suppose}{x}^3+y^{2}z+z^{2}x=3xyz\cdots \left(1\right)$$\frac{x^3+\frac{y^{2}z}{2}+\frac{y^{2}z}{2}+z^{2}x}{4}\ge \sqrt[4]{4\frac{x^3{y}^{2}z{y}^{2}z{z}^{2}x}{4}}$
$\Rightarrow \frac{3xyz}{4}\ge \frac{xyz}{\sqrt{2}}\text{[From (1)]}$
Therefore, $x\ne y\ne z$

(IV) $\frac{x+y+z}{3}\ge \sqrt[3]{3xyz}$
$\Rightarrow (x+y+z{)}^3\ge 27xyz$
$\text{Then,}(x+y+z{)}^3=27xyz\Rightarrow x=y=z$