What's the minimum of $\displaystyle K=x^6+y^6$ if $\displaystyle x-y=1$?
Hi
$\displaystyle K(x)=x^6+(x-1)^6$
$\displaystyle K'(x)=6x^5+6(x-1)^5=6(2x-1)(x^4-2x^3+4x^2-3x+1)$
By successive derivation of $\displaystyle L(x)=x^4-2x^3+4x^2-3x+1$ you can show that L(x) is decreasing up to x=1/2 and then increasing and is always positive
The minimum of K(x) is therefore obtained for x=1/2
Let's try this
If $\displaystyle y \geq 0$ then you can set $\displaystyle y = \sinh^2(\alpha)$ where $\displaystyle \alpha \in [0,+\infty]$
Then $\displaystyle x = 1 + y = \cosh^2(\alpha)$
And $\displaystyle K = \cosh^{12}(\alpha) + \sinh^{12}(\alpha)$
$\displaystyle K'(\alpha) = 12 \cosh(\alpha) \:\sinh(\alpha)\\cosh^{10}(\alpha) + \sinh^{10}(\alpha))$
Therefore $\displaystyle K'(\alpha) \geq 0$
And K(x) is increasing over $\displaystyle [1,+\infty]$
If $\displaystyle y \leq -1$ then you can set $\displaystyle y = -\cosh^2(\alpha)$ where $\displaystyle \alpha \in ]-\infty,0]$
Then $\displaystyle x = 1 + y = -\sinh^2(\alpha)$
And $\displaystyle K = \sinh^{12}(\alpha) + \cosh^{12}(\alpha)$
$\displaystyle K'(\alpha) = 12 \cosh(\alpha) \:\sinh(\alpha)\\cosh^{10}(\alpha) + \sinh^{10}(\alpha))$
Therefore $\displaystyle K'(\alpha) \leq 0$
And K(x) is decreasing over $\displaystyle ]-\infty,0]$
If $\displaystyle -1 \leq y \leq 0$ then you can set $\displaystyle y = -\sin^2(\alpha)$ where $\displaystyle \alpha \in [0,\frac{\pi}{2}]$
Then $\displaystyle x = 1 + y = \cos^2(\alpha)$
And $\displaystyle K = \cos^{12}(\alpha) + \sin^{12}(\alpha)$
$\displaystyle K'(\alpha) = 12 \cos(\alpha) \:\sin(\alpha)\\sin^{10}(\alpha) - \cos^{10}(\alpha))$
I think I've got a simpler solution:
$\displaystyle K'(x)=6x^5+6(x-1)^5=0$
$\displaystyle 6x^5=-6(x-1)^5$
$\displaystyle (\frac{x}{x-1})^5=-1$
$\displaystyle \frac{x}{x-1}=-1$
$\displaystyle x=-x+1$
$\displaystyle 2x=1$
$\displaystyle x=\frac{1}{2}$
Then by using the second derivative test it's easy to prove that it is indeed a min.