About positive real numbers $\displaystyle a, b, c$ we know that $\displaystyle a+b+c=17$ and $\displaystyle a\cdot b\cdot d=64$.

Show that $\displaystyle \max(a,b,c) \geq 8.$

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- May 4th 2010, 10:07 PMArczi1984Real number problem
About positive real numbers $\displaystyle a, b, c$ we know that $\displaystyle a+b+c=17$ and $\displaystyle a\cdot b\cdot d=64$.

Show that $\displaystyle \max(a,b,c) \geq 8.$ - May 4th 2010, 11:04 PMchiph588@
$\displaystyle a+b=17-c $ and $\displaystyle ab=\frac{64}{c} \implies $ WLOG $\displaystyle a=\frac12\left(17-c+\sqrt{(17-c)^2-\frac{256}{c}}\right) $ and $\displaystyle b=\frac12\left(17-c-\sqrt{(17-c)^2-\frac{256}{c}}\right) $

Now assume $\displaystyle a,b,c < 8 $

Consider $\displaystyle f(x)=\frac12\left(17-c+\sqrt{(17-c)^2-\frac{256}{c}}\right)-8 $ on the interval $\displaystyle [1,8) $ and use calculus to find that the minimum is $\displaystyle 0 $ which implies $\displaystyle a\geq8 $ and that gives us a contradiction. - May 4th 2010, 11:17 PMArczi1984
Thanks.

I've question about last statement.

Is the function $\displaystyle f(x)$ constant? On the right side is only 'c' - positive real number (there is no 'x').

And from where we know that $\displaystyle a\geq 8$? - May 5th 2010, 05:28 AMchiph588@
Whoops, it should be $\displaystyle f(x)=\frac12\left(17-x+\sqrt{(17-x)^2-\frac{256}{x}}\right)-8

$.

If the minimum of $\displaystyle f(x) $ is $\displaystyle 0 $, then that means for all $\displaystyle x\in[1,8),\;\;\frac12\left(17-x+\sqrt{(17-x)^2-\frac{256}{x}}\right)\geq8 $ which is a problem for us. - May 5th 2010, 05:31 AMArczi1984
Ok, now it is clear :) Once more thanks for help.