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Question Number 147892 by mathdanisur last updated on 24/Jul/21
ifx;y;z>1then: (x−1)(y−1)(z−1)(x+1)(y+1)(z+1)<xyz8
Answered by mindispower last updated on 24/Jul/21
x−1x+1<x24...?∀x>1 ⇔4(x−1)<x2(x+1) ⇔x3+x2+4−4x>0 ⇔x(x2+1)+4−4x>0 x2+1⩾2x⇒x(x2+1)+4−4x>2x2−4x+4=2(x−1)2+2>0 ⇒∀x>1x−1x+1<x24,y−1y+1<y24,z−1z+1<z24 ⇒(x−1)(y−1)(z−1)(x+1)(y+1)(z+1)<x2y2z264 ⇒(x−1)(y−1)(z−1)(x+1)(y+1)(z+1<xyz8
Commented bymathdanisur last updated on 24/Jul/21
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