1. ## Real 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 $\displaystyle \max(a,b,c) \geq 8.$

2. Originally Posted by Arczi1984
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 $\displaystyle \max(a,b,c) \geq 8.$
$\displaystyle a+b=17-c$ 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.

3. Thanks.
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 $\displaystyle a\geq 8$?

4. Originally Posted by Arczi1984
Thanks.
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 $\displaystyle a\geq 8$?
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.