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| Find a geometric way to prove the inequality! (Posted on 2007-11-07) |
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If 0 < a < b < c < d then prove that
a/b + b/c + c/d + d/a > b/a + c/b + d/c + a/d.
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Submitted by Chesca Ciprian
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Rating: 3.3333 (3 votes)
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Solution:
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Let consider on the graph of the function f(x) = 1/x , 4 point A(a,1/a),B(b,1/b),C(c,1/c),D(d,1/d) with 0 < a < b < c < d. These points form 4 trapezium. The sum of the area of 3 trapezium are smaller than the last trapezium.
So we can write that (b-a)(1/a+1/b) + (c-b)(1/c+1/b) + (d-c)(1/d+1/c) < (d-a)(1/a+1/d).
After calculus we find that :
a/b+b/c+c/d+d/a > b/a+c/b+d/c+a/d!
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