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In the problem a and b are positive real numbers , and two equations of $x^2 + ax + 2b = 0 $ and $x^2 + 2bx + a = 0$.Where $a,b \in \mathbb{R^{+}}$ has real roots. find the minimum value of $(a + b)$.

So far from discriminant property i have,

$a^2\geq8b\\ b^2\geq a\\ a>0\\ b>0$

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    $\begingroup$ This is a nice little problem in that it shows the 'fragility' of the minimum value with respect to the constraints. If you have $a\ge0, b\ge 0$, the minimum is $0$ instead of 6. $\endgroup$ – copper.hat Feb 22 '15 at 18:34
  • $\begingroup$ @copper.hat But in the question it is clearly given that $a$ and $b$ are positive real numbers. $\endgroup$ – R K Feb 22 '15 at 19:18
  • $\begingroup$ I understand that. I was making a pedagogical point, which is that finding minima or infimising values is often a delicate task. For example, it is not clear a priori that a minimum exists in the first place. $\endgroup$ – copper.hat Feb 22 '15 at 19:23
  • $\begingroup$ Uhm.. how do you even arrive at $ a^2 >= 8b$ What discriminant property? You have three variables... $\endgroup$ – dramadeur Mar 19 '15 at 5:58
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First, it should be $a^2\geq 8b$ and $b^2\geq a$.

Hint: From $a^2\geq 8b\geq 0$, deduce $a^4\geq 64b^2\geq64a$.

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  • $\begingroup$ from ur hint , I arrived at $\left(\dfrac{a}{2\sqrt2}\right)^4> b^2>a$ but i m still confused... $\endgroup$ – R K Feb 22 '15 at 17:54
  • $\begingroup$ so $\Large a\geq 4\\\Large and\\ \Large b\geq 2\\\Large (a+b)_{min}=6 ?$ $\endgroup$ – R K Feb 22 '15 at 18:20
  • $\begingroup$ can u confirm that if my answer is correct ? $\endgroup$ – R K Feb 22 '15 at 18:22
  • $\begingroup$ Well, $(a+b)_{\mathrm{min}}=6$ only if $a=4$ and $b=2$ is a valid pair. Is it? (Well, it is, but you have to verify it. Just knowing that $a\geq 4$ and $b\geq 2$ is not enough to assert this, however.) $\endgroup$ – Thomas Andrews Feb 22 '15 at 18:23
  • $\begingroup$ ok, that need to be checked @thomas thanks for notifying that $\endgroup$ – R K Feb 22 '15 at 18:26
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A diagram may help you.

You need to figure out which region is feasible.

The answer is:

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  • $\begingroup$ From the figure $(a+b)_{min} $ seems $0$. so should it be $0$. $\endgroup$ – R K Mar 19 '15 at 10:32
  • $\begingroup$ No. The region should be under the purple curve and above the blue one. $\endgroup$ – Eclipse Sun Mar 19 '15 at 14:44
  • $\begingroup$ So only the red and blue region is valid as in this pic $\endgroup$ – R K Mar 19 '15 at 14:52
  • $\begingroup$ No. The regions you colored is either under the purple curve or above the blue one. Neither satisfies the condition "under the purple curve and above the blue one". $\endgroup$ – Eclipse Sun Mar 19 '15 at 15:19
  • $\begingroup$ ok so i hope u said the yellow region in this pic $\endgroup$ – R K Mar 19 '15 at 18:27

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