If n≥1 a and b are positive real numbers Then prove that: ((a^n + b^n )/(a + b)) ≥ ((a^(n−1) + b^(n−1) )/2)


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Question Number 175815 by Shrinava last updated on 07/Sep/22

$If n ⩾ 1$ $a and b are positive real numbers$ $Then prove that :$ $\frac{a^{n} + b^{n}}{a + b} ⩾ \frac{a^{n - 1} + b^{n - 1}}{2}$
	Answered by mahdipoor last updated on 07/Sep/22
	$⇔ 2a^n +2b^n ≥a^n +b^n +ab^(n−1) +ba^(n−1) ⇔ a^n +b^n ≥ab^(n−1) +ba^(n−1) ⇔a^(n−1) (a−b)≥b^(n−1) (a−b) ⇔ { ((⇔^(a−b≥0) a^(n−1) ≥b^(n−1) ⇔ a≥b)),((⇔^(a−b≤0) a^(n−1) ≤b^(n−1) ⇔ a≤b)) :}$
	$\Leftrightarrow 2 a^{n} + 2 b^{n} ⩾ a^{n} + b^{n} + {a b}^{n - 1} + {b a}^{n - 1}$ $\Leftrightarrow a^{n} + b^{n} ⩾ {a b}^{n - 1} + {b a}^{n - 1}$ $\Leftrightarrow a^{n - 1} (a - b) ⩾ b^{n - 1} (a - b)$ $\Leftrightarrow {\begin{cases} \overset{a - b ⩾ 0}{\Leftrightarrow} a^{n - 1} ⩾ b^{n - 1} \Leftrightarrow a ⩾ b \\ \overset{a - b ⩽ 0}{\Leftrightarrow} a^{n - 1} ⩽ b^{n - 1} \Leftrightarrow a ⩽ b \end{cases}$
	Answered by Cesar1994 last updated on 07/Sep/22

	$(a + b) (a^{n - 1} + b^{n - 1}) = a^{n} + b^{n} + {a b}^{n - 1} + a^{n - 1} b . . . (1)$ $w e p r o o f t h a t a^{n} + b^{n} ⩾ {a b}^{n - 1} + a^{n - 1} b . . . (2)$ $a^{n} - a^{n - 1} b + b^{n} - {a b}^{n - 1} = a^{n - 1} (a - b) + b^{n - 1} (b - a)$ $= (a^{n - 1} - b^{n - 1}) (a - b)$ $without loss of generality, if a ⩾ b > 0$ $\Rightarrow a^{n} - a^{n - 1} b + b^{n} - {a b}^{n - 1} = (a^{n - 1} - b^{n - 1}) (a - b) ⩾ 0$ $\Rightarrow (a + b) (a^{n - 1} + b^{n - 1}) ⩽ 2 (a^{n} + b^{n}), using 1 a n d 2$ $\Rightarrow \frac{a^{n - 1} + b^{n - 1}}{2} ⩽ \frac{a^{n} + b^{n}}{a + b} ◼$
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