To prove 1 2 2 l 1 z z 1 Zz z z Z Z 1 22 2 2 2 z 2 2 z 1 z 1 1 z 0 1 2 1 0 Which is obvious as z 1 z Pe cose 3 sine shortest distance exists along the common normal Slope of normal at P esece 3 cos ece 2 tan 6 1 so cose and sine 3 Hence PE 2 1 A A I a b c a b c bcabca ca bc a b 0 0 1 SOLUTIONS a b2 c ab bc ca ab bc ca 100 ab bc ca a b c ab bc ca 010 ab bc ca ab bc ca a 6 0 0 0 1 100 010 a b c 1 1 and ab bc ca 0 2 Now a b c a b c a b c ab bc ca 3abc a b c 3 3 Now a b c a b 2 ab bc ca 1 2 0 1 a b c 1 Now from 3 a b c 1 3 4 a b c 4 1 2 1 2 4 C 4C Alternate A A 1 AA A 1 a b c 3abc 1 a b c 3abc 1 since a b c are positive real number a b c 2 3abc from AM2 GM since a b c are real positive number 1 1 2 n k k r k r x y 7 n r n n r r n k k r