If x, y, z be positive real numbers such that $x^{2} + y^{2} + z^{2} = 27$ , then show that $x^{3} + y^{3} + z^{2} \geq 81$

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Solution

Applying Cauchy- Schwarz inequality to the two sets of numbers
$x^{\frac{3}{2}}, y^{\frac{3}{2}}, z^{\frac{3}{2}};$
$x^{\frac{1}{2}}, y^{\frac{1}{2}}, z^{\frac{1}{2}}$
we have
${(x^{\frac{3}{2}} . x^{\frac{1}{2}} + y^{\frac{3}{2}} . y^{\frac{1}{2}} + z^{\frac{3}{2}} . z^{\frac{1}{2}})}^{2} \leq (x^{3} + y^{3} + z^{3}) (x + y + z)$
$(x^{2} + y^{2} + z^{2})^{2} \leq (x^{3} + y^{3} + z^{3}) (x + y + z)$ ...........(1)
Applying Cauchy Schwarz inequality to the two sets numbers x, y, z ; 1, 1, 1
$(x + y + z)^{2} \leq (x^{2} + y^{2} + z^{2}) (1 + 1 + 1$ ).........(2)
Squaring (1)
$(x^{2} + y^{2} + z^{2})^{4} \leq (x^{3} + y^{2} + z^{3})^{2} (x + y + z)^{2}$
Combining (3) and (2)
$(27)^{4} \leq (x^{3} + y^{2} + z^{3})^{2} (x + y + z)^{2} .3$
$(27)^{4} \leq (x^{3} + y^{2} + z^{3})^{2} . 27.3$
$3^{8} \leq (x^{3} + y^{2} + z^{3})^{2}$
$\Rightarrow (x^{3} + y^{2} + z^{3})^{2} \geq (3^{4})^{2}$
$x^{3} + y^{3} + z^{3} \geq 81$ .

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