We have 0<x<a and m,n ∈N. Prove x^m (a−x)^n ≤ ((m^m n^n )/((m+n)^(m+n) ))∙a^(m+n)


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Question Number 182791 by Matica last updated on 14/Dec/22

$W e h a v e 0 < x < a a n d m, n \in N .$ $P r o v e x^{m} {(a - x)}^{n} ⩽ \frac{m^{m} n^{n}}{{(m + n)}^{m + n}} \cdot a^{m + n}$
	Answered by mahdipoor last updated on 14/Dec/22

	$g e t f (x) = x^{m} {(a - x)}^{n} 0 < x < a$ $\frac{d f}{d x} = {m x}^{m - 1} {(a - x)}^{n} - n {(a - x)}^{n - 1} x^{m} = 0$ $= {(a - x)}^{n - 1} x^{m - 1} (m (a - x) - n x) \Rightarrow x = \frac{m a}{m + n}$ $m a x f = f (\frac{m a}{m + n}) = \frac{m^{m} n^{n} a^{m + n}}{{(m + n)}^{m + n}}$ $f ⩽ m a x f \Rightarrow x^{m} {(a - x)}^{n} ⩽ \frac{m^{m} n^{n} a^{m + n}}{{(m + n)}^{m + n}}$
	Commented byMatica last updated on 15/Dec/22

	$t h a n k y o u a l o t$
	Answered by dre23 last updated on 15/Dec/22
	$x^m (1−x)^n ≤((m^m n^n )/((m+n)^(m+n) )),....S by x→ax⇔∀ 0<x<1 .....S ⇔((x/m))^m (((1−x)/n))^n ≤(1/((n+m)^(n+m) ))⇔S mln((x/m))+nln(((1−x)/n))....S x→^f ln(x) is concave,f′′=−(1/x^2 )<0,∀x∈]0,1] (m/(n+m))ln((x/m))+(n/(n+m))ln(((1−x)/n))≤ln((1/(n+m))) (n+m){(m/(m+n))ln((x/m))+(n/(n+m))ln(((1−x)/n))}≤ (n+m)ln((x/(n+m))+((1−x)/(n+m)))≤(n+m)ln((1/(n+m)))=ln((1/((n+m)^(n+m) ))) ⇔s≤ln((1/((n+m)^(n+m) ))) tack e ⇔((x/m))^m (((1−x)/n))^n ≤((1/(n+m)))^(n+m) ..True$
	$x^{m} {(1 - x)}^{n} ⩽ \frac{m^{m} n^{n}}{{(m + n)}^{m + n}}, . . . . S$ $b y x \to a x \Leftrightarrow \forall 0 < x < 1 . . . . . S$ $\Leftrightarrow {(\frac{x}{m})}^{m} {(\frac{1 - x}{n})}^{n} ⩽ \frac{1}{{(n + m)}^{n + m}} \Leftrightarrow S$ $m l n (\frac{x}{m}) + n l n (\frac{1 - x}{n}) . . . . S$ $x \overset{f}{\to} l n (x) i s c o n c a v e, f^{″} = - \frac{1}{x^{2}} < 0, \forall x \in] 0, 1]$ $\frac{m}{n + m} l n (\frac{x}{m}) + \frac{n}{n + m} l n (\frac{1 - x}{n}) ⩽ l n (\frac{1}{n + m})$ $(n + m) {\frac{m}{m + n} l n (\frac{x}{m}) + \frac{n}{n + m} l n (\frac{1 - x}{n})} ⩽$ $(n + m) l n (\frac{x}{n + m} + \frac{1 - x}{n + m}) ⩽ (n + m) l n (\frac{1}{n + m}) = l n (\frac{1}{{(n + m)}^{n + m}})$ $\Leftrightarrow s ⩽ l n (\frac{1}{{(n + m)}^{n + m}}) t a c k e$ $\Leftrightarrow {(\frac{x}{m})}^{m} {(\frac{1 - x}{n})}^{n} ⩽ {(\frac{1}{n + m})}^{n + m} . . T r u e$
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