# In the problem find the minimum value of $(a + b)$.

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,

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

• 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. – copper.hat Feb 22 '15 at 18:34
• @copper.hat But in the question it is clearly given that $a$ and $b$ are positive real numbers. – R K Feb 22 '15 at 19:18
• 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. – copper.hat Feb 22 '15 at 19:23
• Uhm.. how do you even arrive at $a^2 >= 8b$ What discriminant property? You have three variables... – dramadeur Mar 19 '15 at 5:58

First, it should be $a^2\geq 8b$ and $b^2\geq a$.

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

• from ur hint , I arrived at $\left(\dfrac{a}{2\sqrt2}\right)^4> b^2>a$ but i m still confused... – R K Feb 22 '15 at 17:54
• so $\Large a\geq 4\\\Large and\\ \Large b\geq 2\\\Large (a+b)_{min}=6 ?$ – R K Feb 22 '15 at 18:20
• can u confirm that if my answer is correct ? – R K Feb 22 '15 at 18:22
• 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.) – Thomas Andrews Feb 22 '15 at 18:23
• ok, that need to be checked @thomas thanks for notifying that – R K Feb 22 '15 at 18:26

• From the figure $(a+b)_{min}$ seems $0$. so should it be $0$. – R K Mar 19 '15 at 10:32