Compare-1900-3-4-99-3-4-and-1999-3-4-

Question Number 149469 by mathdanisur last updated on 05/Aug/21

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1900^{\frac{3}{4}} + 99^{\frac{3}{4}} and 1999^{\frac{3}{4}}

Answered by mindispower last updated on 05/Aug/21

f (x) = x^{\frac{3}{4}} + {(1 - x)}^{\frac{3}{4}}, x \in [0, 1 [, f (1 - x) = f (x)

f^{'} (x) = \frac{3}{4} (\frac{1}{x^{\frac{1}{4}}} - \frac{1}{{(1 - x)}^{\frac{1}{4}}}) = \frac{3}{4} (\frac{{(1 - x)}^{\frac{1}{4}} - x^{\frac{1}{4}}}{{(x (1 - x))}^{\frac{1}{4}}})

i f x ⩽ \frac{1}{2}, f^{'} > 0

i f x \in] \frac{1}{2}, 1 [f^{'} < 0

f (1) = f (0) = 1

\Rightarrow \forall x \in [0, 1] f (x) ⩾ 1

f o r x = \frac{1900}{1999}

f (\frac{1900}{1999}) = {(\frac{1900}{1999})}^{\frac{3}{4}} + {(1 - \frac{1900}{1999})}^{\frac{3}{4}} ⩾ 1

\Rightarrow {(1900)}^{\frac{3}{4}} + 99^{\frac{3}{4}} ⩾ 1999^{\frac{3}{4}}

Commented by mathdanisur last updated on 05/Aug/21

Thank You Ser Cool

Commented by mathdanisur last updated on 06/Aug/21

Ser, can it be written as {Newton}^{'} s

Binomial or inequality . ?