if a>0 b>0 a≤b show that a^2 ≤(((2ab)/(a+b)))^2 ≤ab≤(((a+b)/2))^2 ≤((a^2 +b^2 )/2)≤b^2


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Question Number 82667 by M±th+et£s last updated on 23/Feb/20

$i f a > 0 b > 0 a ⩽ b$ $s h o w t h a t$ $a^{2} ⩽ {(\frac{2 a b}{a + b})}^{2} ⩽ a b ⩽ {(\frac{a + b}{2})}^{2} ⩽ \frac{a^{2} + b^{2}}{2} ⩽ b^{2}$
	Answered by TANMAY PANACEA last updated on 23/Feb/20
	$b^2 −((a^2 +b^2 )/2) (((b+a)(b−a))/2)>0 so b^2 >((a^2 +b^2 )/2) [equality sivn if a=b] ((a^2 +b^2 )/2)−(((a+b)/2))^2 =((2a^2 +2b^2 −a^2 −b^2 −2ab)/4)→(((b−a)^2 )/2^2 )>0 so ((a^2 +b^2 )/2)>(((a+b)/2))^2 (((a+b)/2))^2 −ab =(((a−b)^2 )/2^2 ) so (((a+b)/2))^2 >ab ab−(((2ab)/(a+b)))^2 =ab×((a^2 +2ab+b^2 −4ab)/((a+b)^2 ))→ab×(((a−b)/(a+b)))^2 so ab>(((2ab)/(a+b)))^2 (((2ab)/(a+b)))^2 −a^2 =a^2 ×((4b^2 −a^2 −2ab−b^2 )/((a+b)^2 )) (a^2 /((a+b)^2 ))×(3b^2 −3ab+ab−a^2 ) (a^2 /((a+b)^2 ))×{3b(b−a)+a(b−a)} (a^2 /((a+b)^2 ))×(b−a)(3b+a)>0 since b>a$
	$b^{2} - \frac{a^{2} + b^{2}}{2}$ $\frac{(b + a) (b - a)}{2} > 0 s o b^{2} > \frac{a^{2} + b^{2}}{2} [e q u a l i t y s i v n i f a = b]$ $\frac{a^{2} + b^{2}}{2} - {(\frac{a + b}{2})}^{2}$ $= \frac{2 a^{2} + 2 b^{2} - a^{2} - b^{2} - 2 a b}{4} \to \frac{{(b - a)}^{2}}{2^{2}} > 0$ $s o \frac{a^{2} + b^{2}}{2} > {(\frac{a + b}{2})}^{2}$ ${(\frac{a + b}{2})}^{2} - a b$ $= \frac{{(a - b)}^{2}}{2^{2}} s o {(\frac{a + b}{2})}^{2} > a b$ $a b - {(\frac{2 a b}{a + b})}^{2}$ $= a b \times \frac{a^{2} + 2 a b + b^{2} - 4 a b}{{(a + b)}^{2}} \to a b \times {(\frac{a - b}{a + b})}^{2}$ $s o a b > {(\frac{2 a b}{a + b})}^{2}$ ${(\frac{2 a b}{a + b})}^{2} - a^{2}$ $= a^{2} \times \frac{4 b^{2} - a^{2} - 2 a b - b^{2}}{{(a + b)}^{2}}$ $\frac{a^{2}}{{(a + b)}^{2}} \times (3 b^{2} - 3 a b + a b - a^{2})$ $\frac{a^{2}}{{(a + b)}^{2}} \times {3 b (b - a) + a (b - a)}$ $\frac{a^{2}}{{(a + b)}^{2}} \times (b - a) (3 b + a) > 0 s i n c e b > a$
	Commented by M±th+et£s last updated on 23/Feb/20

	$g o d b l e s s y o u s i r$
	Commented by TANMAY PANACEA last updated on 23/Feb/20

	$b l e s s i n g s h o s e r t o a l l$
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