if x>0 , r=pq , 1≤p≤q then: 1+rx ≤ (1+qx)^p ≤ (1+px)^q ≤ (1+x)^r


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Question Number 144095 by mathdanisur last updated on 21/Jun/21

$i f x > 0, r = p q, 1 ⩽ p ⩽ q t h e n :$ $1 + r x ⩽ {(1 + q x)}^{p} ⩽ {(1 + p x)}^{q} ⩽ {(1 + x)}^{r}$
	Answered by mitica last updated on 21/Jun/21

	${(1 + x)}^{r} = {[{(1 + x)}^{p}]}^{q} \overset{b e r n o u l l i}{⩾} {(1 + p x)}^{q}$ ${(1 + q x)}^{p} \overset{b e r n o u l l i}{⩾} 1 + p q x = 1 + r x$ $f (t) = \frac{l n (1 + t)}{t} ↘$ $p x ⩽ q x \Rightarrow \frac{l n (1 + p x)}{p} ⩾ \frac{l n (1 + q x)}{q} \Rightarrow {(1 + p x)}^{q} ⩾ {(1 + q x)}^{p}$
	Commented bymathdanisur last updated on 21/Jun/21

	$a l o t p e r f e c t S i r t h a n k y o u$
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