The least natural number a for which $x + a x^{- 2} > 2 f o r a l l x ϵ R^{+}$

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Solution

x + $\frac{a}{x^{2}} > 2 \forall x ϵ R^{+} x^{3} - 2 x^{2} + a > 0$
f(x) = $x^{3} - 2 x^{2} + a$
f'(x) = $3 x^{2} - 4 x = x (3 x - 4)$
f $(\frac{4}{3}) > 0 {(\frac{4}{3})}^{3} - 2 {(\frac{4}{3})}^{2} + a > 0$
$\frac{64}{27} - \frac{32}{9} + a > 0$
$\frac{64 - 96}{27} + a > 0$
$a > \frac{32}{27}$ least natural number is 2.

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