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
Answer & Explanation
Solved by verified expert
Get Answers to Unlimited Questions
Join us to gain access to millions of questions and expert answers. Enjoy exclusive benefits tailored just for you!
Membership Benefits:
Unlimited Question Access with detailed Answers
Zin AI - 3 Million Words
10 Dall-E 3 Images
20 Plot Generations
Conversation with Dialogue Memory
No Ads, Ever!
Access to Our Best AI Platform: Flex AI - Your personal assistant for all your inquiries!
