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Show that the minimum value of $\frac {(x+a)(x+b)}{(x+c)}$, where a$\gt$c, b$\gt$c, is $(\sqrt{a-c}+\sqrt{b-c})^{2}$ for real values of x$\gt-c$.

I did $$\frac {(x+a)(x+b)}{(x+c)}=y$$ and then took its discriminant greater than zero. This led me to $$y^2-2(a+b-2c)y+(a-b)^2\gt0$$

I also tried differentiating the expression as follows. $$y'= \frac {(x+c)[2x+(a+b)]-[x^2+(a+b)x+ab]}{(x+c)^2}=0$$ $$\therefore x^2+2cx+(a+b)c-ab=0$$ I am unable to proceed after this. Please help.

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    $\begingroup$ I'm not sure how you got from your expression for $y$ to your final equation. Are you allowed to use calculus? The standard approach would be to differentiate the original expression with respect to $x$, set it equal to zero, clear out any denominators, and then find the roots of teh resulting polynomial in $x$. $\endgroup$
    – user7530
    Nov 1, 2013 at 5:16
  • $\begingroup$ Can you show me the steps? I did try to differentiate it, but I think I'm going wrong somewhere. $\endgroup$
    – Tejas
    Nov 1, 2013 at 5:26

2 Answers 2

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Your last equation is correct. Solving it for $x$ gives you the two extrema of your function; say that the roots of the quadratic equation (your last one) are $x_1$ and $x_2$. Because of the signs, $x_1$ corresponds to the maximum and $x_2$ to the minimum of the function. Compute now the corresponding value $y_2$ (you will need to work for simplifying them). From what I got, $$y_2 = (a + b - 2 c) + 2 \sqrt{(a-c) (b-c)}$$ Manipulating $y_2$ shows that it is equal to $$(\sqrt{a-c} + \sqrt{b-c})^2$$ I hope and wish this helps you to continue.

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  • $\begingroup$ Did you find x1 and x2 by applying Quadratic Formula to the last equation? $\endgroup$
    – Tejas
    Nov 1, 2013 at 12:43
  • $\begingroup$ Yes, for sure. I do not think there is another way. If you find any (or you are told), please let me know. $\endgroup$ Nov 1, 2013 at 13:19
  • $\begingroup$ No, I too think that's the only way. $\endgroup$
    – Tejas
    Nov 1, 2013 at 13:52
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I'm going to avoid using calculus to solve this question. Instead, I have a different approach which uses simple algebra. Let's call this given expression $z$. Put $y = c + x$ $$⇒z = \frac {(y-c+a)(y-c+b)}{y}$$ Simplifying this by opening the brackets we get, $$⇒z = \frac {(a-c)(b-c) + y(a-c) + y(b-c) + y^2}{y}$$ $$⇒z = \frac {(a-c)(b-c)}{y} + (a-c) + (b-c) + y$$ Now, here we do a bit of manipulation. We add and subtract $2 \sqrt{(a-c) (b-c)}$. Why? Keep looking. $$⇒z = \frac {(a-c)(b-c)}{y} + (a-c) + (b-c) + y + 2 \sqrt{(a-c) (b-c)} - 2 \sqrt{(a-c) (b-c)}$$ Here, we club $\frac {(a-c)(b-c)}{y}, y$ and $2 \sqrt{(a-c) (b-c)}$ to get a squared term. That's why I added and subtacted $2 \sqrt{(a-c) (b-c)}$. $$⇒z = \frac {(a-c)(b-c)}{y} + y + 2 \sqrt{(a-c) (b-c)} + (a-c) + (b-c) - 2 \sqrt{(a-c) (b-c)}$$ $$⇒ z = (\frac {\sqrt{(a-c)(b-c)}}{\sqrt{y}} - \sqrt{y})^2+ (a+b-2c) - 2 \sqrt{(a-c) (b-c)}$$ Now, since $$⇒(\frac {\sqrt{(a-c)(b-c)}}{\sqrt{y}} - \sqrt{y})^2 ≥ 0$$ This is because it is a squared term. It can never be negative, it is only either positive or at the minimum, zero. So,the expression $z$ is minimum only when $$⇒(\frac {\sqrt{(a-c)(b-c)}}{\sqrt{y}} - \sqrt{y})^2 = 0$$ Hence, the minimum value of $z$ is, i.e, z reduces to $$⇒ z = (a+b-2c) - 2 \sqrt{(a-c) (b-c)}$$ Which upon simplification is, $$⇒ z = (\sqrt{a-c} + \sqrt{b-c})^2$$ Which is the required minimum value.

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