Suppose that a, b, A, B are all > 0. Is it always true that (a+b)/(A+B) <= (a/A)+(b/B)?

Question

Please provide a proof, or a specific counterexample with

a,b,A,B being four specific numbers. Does Jensen's finite inequality apply in some way to this problem?

http://en.wikipedia.org/wiki/Jensen

Answer 1

(a+b)/(A+B) == a/(A+B)+ b/(A+B)

now  (A+B)&gt;A ===&gt; 1/(A+B)&lt;1/A then:

a/(A+B)&lt;a/A (a&gt;0 for hp) similary:

b/(A+B)&lt;b/B

then:

(a+b)/(A+B) == a/(A+B)+ b/(A+B)&lt;= (a/A)+(b/B)

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Edit

yes it is true, I wrote wrong:

we have a large inequality..

(a+b)/(A+B) &lt; (a/A)+(b/B)

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very excellent allen s

Answer 2

Almost true, not exactly true because they cannot ever be equal.

Proof (by negation/contradiction):

(1) Assume the proposition is false, i.e., that

(a + b)/(A + B) &gt;= (a/A) + (b/B)

(2) Then, multiply both sides by the positive quantity

(A + B), resulting in the following relation:

a + b &gt;= (A + B)*(a/A) + (A + B)*(b/B)

(3) Expanding the right side, we have

a + b &gt;= a + B*a/A + A*b/B + b = (a + b) + (B*a/A + A*b/B)

(4) The latter relation implies that (B*a/A + A*b/B) &lt;= 0.

Since all the parameters therein are non-zero and positive,

this relation cannot be true. Thus, the original

premise/assumption must be rejected and the

opposite must be true always, meaning that

(a + b)/(A + B) &lt; (a/A) + (b/B)