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Question Number 147892 by mathdanisur last updated on 24/Jul/21

if   x;y;z>1   then:  (√((((x−1)(y−1)(z−1))/((x+1)(y+1)(z+1))) )) < ((xyz)/8)

ifx;y;z>1then: (x1)(y1)(z1)(x+1)(y+1)(z+1)<xyz8

Answered by mindispower last updated on 24/Jul/21

((x−1)/(x+1))<(x^2 /4)...? ∀x>1  ⇔4(x−1)<x^2 (x+1)  ⇔x^3 +x^2 +4−4x>0  ⇔x(x^2 +1)+4−4x>0  x^2 +1≥2x⇒x(x^2 +1)+4−4x>2x^2 −4x+4=2(x−1)^2 +2>0  ⇒∀x>1 ((x−1)/(x+1))<(x^2 /4),((y−1)/(y+1))<(y^2 /4),((z−1)/(z+1))<(z^2 /4)  ⇒(((x−1)(y−1)(z−1))/((x+1)(y+1)(z+1)))<((x^2 y^2 z^2 )/(64))  ⇒(√(((x−1)(y−1)(z−1))/((x+1)(y+1)(z+1)))<((xyz)/8)

x1x+1<x24...?x>1 4(x1)<x2(x+1) x3+x2+44x>0 x(x2+1)+44x>0 x2+12xx(x2+1)+44x>2x24x+4=2(x1)2+2>0 x>1x1x+1<x24,y1y+1<y24,z1z+1<z24 (x1)(y1)(z1)(x+1)(y+1)(z+1)<x2y2z264 (x1)(y1)(z1)(x+1)(y+1)(z+1<xyz8

Commented bymathdanisur last updated on 24/Jul/21

Thankyou Sir cool

ThankyouSircool

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