Prove the inequality $(x^4+y^4+z^4)+(x^5+y^5+z^5)+(x-y)^6+(y-z)^6+(z-x)^6 \le6$ Let $x,y,z \in [0;1]$. Prove the inequality
$$(x^4+y^4+z^4)+(x^5+y^5+z^5)+(x-y)^6+(y-z)^6+(z-x)^6 \le6$$
My work so far:
Let $f(x)=(x^4+y^4+z^4)+(x^5+y^5+z^5)+(x-y)^6+(y-z)^6+(z-x)^6 -6$, where $x\in[0;1]$
$f(0)=(y^4+z^4)+(y^5+z^5)+(y)^6+(y-z)^6+(z)^6 -6$
 A: @Alex Silva starts off with the reduction
$$\begin{align*} & (x^4+y^4+z^4)+(x^5+y^5+z^5)+(x-y)^6+(y-z)^6+(z-x)^6 \leq \\& \leq 2(x^2+y^2+z^2)+(x-y)^2+(y-z)^2+(z-x)^2\end{align*}$$
Now note that
$$(x - y)^2 \leq (1 - x)^2 + (1 - y)^2$$
which is true because it is equivalent to
$$-2xy \leq -2x + 1 - 2y + 1 \implies x + y \leq 1 + xy \implies y(1 - x) \leq 1 - x$$
The last inequality is certainly true because $(1 - x) \geq 0$ and $ y \in [0,1]$. From here, note that
$$(1 - x)^2 \leq 1 - x^2 \qquad x \in [0,1]$$
So we get
$$(x - y)^2 \leq 2 - x^2 - y^2$$
We can repeat this argument for the other terms to get
$$(x - y)^2 + (x - z)^2 + (y - z)^2 \leq 6 - 2(x^2 + y^2 + z^2)$$
$$\implies 2(x^2 + y^2 + z^2) + (x - y)^2 + (x - z)^2 + (y - x)^2 \leq 6$$
Again, credits to Alex Silva for giving the reduction. Let me know if there are any algebra mistakes.
EDIT:
In general, we have for $x,y \in [0,1]$ the following fact
$$|x - y|^k \leq (1 - x)^k + (1 - y)^k \leq (1 - x^k) + (1 - y^k)$$
where the first inequality follows from the fact that given $x$ and $y$, the distance between them is always less than the distance between $x$ and $1$ and/or the distance between $y$ and $1$. The second inequality follows from the fact that
$$(1 - x^k) = (1 - x)(1 + x + \dots + x^{k-1}) \geq (1 - x)$$
$$\implies (1 - x^k) \geq (1 - x) \geq (1 - x)^k$$
so really the proof is unnecessary because of this general case.
