Question

If $x,y,z\in +R$ and $x^2+y^2+z^2=27$, then $x^3+y^3+z^3$ has

• A. Minimum value of $81$
• B. Maximum value of $81$
• C. Maximum value of $27$
• D. Minimum value of $27$

Answer 1

The correct option is A Minimum value of $81$

Given that $x^2+y^2+z^2=27$

We know that $A.M.\ge G.M.$

Therefore, $\frac{x^2+y^2+z^2}{3}\ge \sqrt[3]{3x^2{y}^2{z}^2}$

$⟹\frac{27}{3}\ge \sqrt[3]{3x^2{y}^2{z}^2}$

$⟹9\ge (xyz{)}^{\frac{2}{3}}$

$⟹(3^2{)}^{\frac{3}{2}}\ge xyz$

$⟹3^3\ge xyz$

$⟹xyz\le 27$ ————-(1)

Similarly, $\frac{x^3+y^3+z^3}{3}\ge \sqrt[3]{3x^3{y}^3{z}^3}$

$⟹\frac{x^3+y^3+z^3}{3}\ge xyz$

$⟹x^3+y^3+z^3\ge 3xyz$ ————-(2)

From (1) $xyz$ has a macimum value of 27. Substituting $xyz=27$ we get

$⟹x^3+y^3+z^3\ge 3\ast 27$

$⟹x^3+y^3+z^3\ge 81$

Therefore, $x^3+y^3+z^3$ has minimum value of 81.