Real number problem

**Arczi1984** · May 4th 2010, 10:07 PM

About positive real numbers $a, b, c$ we know that $a+b+c=17$ and $a\cdot b\cdot d=64$ .
Show that $\max(a,b,c) \geq 8.$

**chiph588@** · May 4th 2010, 11:04 PM

Originally Posted by Arczi1984

About positive real numbers $a, b, c$ we know that $a+b+c=17$ and $a\cdot b\cdot d=64$ .
Show that $\max(a,b,c) \geq 8.$

$a+b=17-c$ and $ab=\frac{64}{c} \implies$ WLOG $a=\frac12\left(17-c+\sqrt{(17-c)^2-\frac{256}{c}}\right)$ and $b=\frac12\left(17-c-\sqrt{(17-c)^2-\frac{256}{c}}\right)$

Now assume $a,b,c < 8$

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

**Arczi1984** · May 4th 2010, 11:17 PM

Thanks.
I've question about last statement.
Is the function $f(x)$ constant? On the right side is only 'c' - positive real number (there is no 'x').
And from where we know that $a\geq 8$ ?

**chiph588@** · May 5th 2010, 05:28 AM

Originally Posted by Arczi1984

Thanks.
I've question about last statement.
Is the function $f(x)$ constant? On the right side is only 'c' - positive real number (there is no 'x').
And from where we know that $a\geq 8$ ?

Whoops, it should be $f(x)=\frac12\left(17-x+\sqrt{(17-x)^2-\frac{256}{x}}\right)-8<br />$ .

If the minimum of $f(x)$ is $0$ , then that means for all $x\in[1,8),\;\;\frac12\left(17-x+\sqrt{(17-x)^2-\frac{256}{x}}\right)\geq8$ which is a problem for us.

**Arczi1984** · May 5th 2010, 05:31 AM

Ok, now it is clear

Once more thanks for help.

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