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.
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.