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$a>0,\; b>0,\; S=$parameter $>0$. $$a+b+\dfrac{S-2}{2(a+b)} \longrightarrow min$$ With condition that $a\cdot b =1$

Using inequality of arithmetic and geometric means we get:

$$a+b+\dfrac{S-2}{2(a+b)} \geq 2+ \dfrac{S-2}{2(a+b)}$$ With equality when $a=b$. This is true, but the answer to this problem is that if $S\leq 10$ this will indeed be the answer, but if $S>10$, $a=b$ no longer gives the best result.

I am mainly interested in relatively "simple" and short solutions and methods of this problem.

Karlis Olte
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  • Have you tried Lagrange's approach to get a general solution yet? – AlexR Oct 15 '13 at 09:13
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    I suggest you replace "b" by "1/a" (this is your condition). Then you have a function of "a" and "S". Compute its derivative and set it equal to zero. This will give you a relation between "a" and "S". You will proably find radicals of quadratics in "S".Does this help ? – Claude Leibovici Oct 15 '13 at 09:15

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For $S\ge 10$, you can use AM-GM like this: $$(a+b)+\dfrac{S-2}{2(a+b)} \ge 2\sqrt{\frac{S-2}{2}}$$ with equality iff $2(a+b)^2=S-2$. From this and $ab = 1$, you can solve for $a, b$.

Macavity
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  • Perfect! But I have a philosophical question: ok, this is the answer... but is this a solution? Well, look..there already was an inequality that was true for all $S>0$ and now there is a better inequality for $S\geq10$, do we know there isn't even better one for some $S$? – Karlis Olte Oct 15 '13 at 10:02
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    The key here is that it cannot be better, because we have found a minimum independent of $(a, b)$, which is in fact achieved for some value of $(a, b)$. – Macavity Oct 15 '13 at 10:10