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Question Number 153676 by SANOGO last updated on 09/Sep/21

	Answered by som(math1967) last updated on 09/Sep/21

	$1 . a > 0, b > 0$ ${(a - b)}^{2} ⩾ 0 [i f a = b t h e n (a - b) = 0]$ $\Rightarrow {(a + b)}^{2} - 4 a b ⩾ 0$ $\Rightarrow {(\frac{a + b}{2})}^{2} ⩾ a b$ $∴ \frac{a + b}{2} ⩾ \sqrt{a b}$
	Commented by SANOGO last updated on 09/Sep/21

	$b i e n / m a i s s t p q u e l e s t l e s u p e r i e u r e t l^{'} i n f e r i e u r$
	Commented by som(math1967) last updated on 09/Sep/21

	$j e n e c o n n a i s p a s l e f r a n c a i s$
	Answered by puissant last updated on 11/Sep/21
	$1) ((√a)−(√b))^2 ≥0 ⇒ a−2(√(ab))+b≥0 ⇒ a+b≥2(√(ab)) ⇒ ((a+b)/2)≥(√(ab)). 2) ..... 3) A={(m/n)+((4n)/m): n,m∈N^∗ } A={0, 5, ((17)/2), 4, ((20)/3), ....} →SupA n′existe pas.. → InfA=minA=0..$
	$1) {(\sqrt{a} - \sqrt{b})}^{2} ⩾ 0$ $\Rightarrow a - 2 \sqrt{a b} + b ⩾ 0$ $\Rightarrow a + b ⩾ 2 \sqrt{a b}$ $\Rightarrow \frac{a + b}{2} ⩾ \sqrt{a b} .$ $2)$ $. . . . .$ $3)$ $A = {\frac{m}{n} + \frac{4 n}{m} : n, m \in N^{*}}$ $A = {0, 5, \frac{17}{2}, 4, \frac{20}{3}, . . . .}$ $\to S u p A n^{'} e x i s t e p a s . .$ $\to I n f A = m i n A = 0 . .$
	Commented by SANOGO last updated on 09/Sep/21

	$o k m e r c i b i e n l e t o u t p u i s s a n t$
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