I came up with a simple algebraic problem, which I have been spending sometime to prove it.

Let $a,b,c,d$ be positive real numbers and $w_1 + w_2 = 1$ where both $w_1$ and $w_2$ are strictly positive. We know that the values chosen for $a, b, c, d$ should satisfy the following inequality, $$\frac{a}{b} > \frac{c}{d}$$

Now I want to show that there exists $w_1$ and $w_2$ pairs such that,

$$\frac{a}{b} \ge \frac{w_1 a + w_2 c }{w_1 b + w_2 d}$$

will also held. I feel like the proof should be quiet simple, but I couldn't come up with satisfying way to prove this. Any help would be appreciated.


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    $\begingroup$ What do you mean by "$a/b>c/d$ holds for all values of $a,b,c,d$"? This is a false premise, whence anything can be deduced from it. If you want to say $a/b>c/d$ holds for a particular tuple $(a,b,c,d)$, then your inquality $$\frac{a}{b}\geq \frac{w_1a+w_2c}{w_1b+w_2d}$$ holds for all $w_1,w_2\geq 0$ such that not both are zero (and the equality occurs iff $w_2=0$). $\endgroup$ – Batominovski Aug 9 '15 at 16:27
  • $\begingroup$ Thanks @Batominovski you are right, I fixed the post. But I still couldn't get the proof. $\endgroup$ – mathamania Aug 9 '15 at 16:36

Hint: Look at $$\frac{a}{b}-\frac{w_1a+w_2c}{w_1b+w_2d},$$ bring to a common denominator, and simplify the numerator.

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    $\begingroup$ Which it is, since from $\frac{a}{b}\gt \frac{c}{d}$ we can conclude by multiplying through by $bd$ (positive) that $ad\gt bc$, and therefore $w_2(ad-bc)\gt 0$. $\endgroup$ – André Nicolas Aug 9 '15 at 19:38
  • $\begingroup$ Oh thanks @Andre that was so simple, now I feel like a dumb :). $\endgroup$ – mathamania Aug 9 '15 at 19:40
  • $\begingroup$ You are welcome. As long as you now see things are simple, everything is fine. A fairly common strategy to prove that $x\gt y$ is to show that $x-y$ is positive, $\endgroup$ – André Nicolas Aug 9 '15 at 20:38

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