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Question Number 82667 by M±th+et£s last updated on 23/Feb/20
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
ifa>0b>0a⩽bshowthata2⩽(2aba+b)2⩽ab⩽(a+b2)2⩽a2+b22⩽b2
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
b2−a2+b22(b+a)(b−a)2>0sob2>a2+b22[equalitysivnifa=b]a2+b22−(a+b2)2=2a2+2b2−a2−b2−2ab4→(b−a)222>0soa2+b22>(a+b2)2(a+b2)2−ab=(a−b)222so(a+b2)2>abab−(2aba+b)2=ab×a2+2ab+b2−4ab(a+b)2→ab×(a−ba+b)2soab>(2aba+b)2(2aba+b)2−a2=a2×4b2−a2−2ab−b2(a+b)2a2(a+b)2×(3b2−3ab+ab−a2)a2(a+b)2×{3b(b−a)+a(b−a)}a2(a+b)2×(b−a)(3b+a)>0sinceb>a
Commented by M±th+et£s last updated on 23/Feb/20
god bless you sir
godblessyousir
Commented by TANMAY PANACEA last updated on 23/Feb/20
blessing shoser to all
blessingshosertoall

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