Show: $(x+y)^4 \leq 8(x^4 + y^4)$ I wish to show the following statement:
$
\forall x,y \in \mathbb{R}
$
$$
(x+y)^4 \leq 8(x^4 + y^4) 
$$
What is the scope for generalisaion?
Edit:
Apparently the above inequality can be shown using the Cauchy-Schwarz inequality. Could someone please elaborate,  stating the vectors you are using in the Cauchy-Schwarz inequality: 
$\ \ \forall \ \ v,w \in V, $ an inner product space,
$$|\langle v,w\rangle|^2 \leq \langle v,v \rangle \cdot \langle w,w \rangle$$
where $\langle v,w\rangle$ is an inner product.
 A: Apply Cauchy-Schwarz inequality twice: $x^4 + y^4 \geq \dfrac{1}{2}\left(x^2+y^2\right)^2 \geq \dfrac{1}{2}\left(\dfrac{1}{2}\left(x+y\right)^2\right)^2 = \dfrac{1}{8}\left(x+y\right)^4$.
A: A more general result is ($x,y\geq 0$, $p\geq 1$)
$$(x+y)^p \leq 2^{p-1} (x^p+y^p),$$
which is direct consequence of convexity of $t\mapsto t^p$.
A: If you instead consider $$\left( \frac{x}{2} + \frac{y}{2} \right)^4$$ we know that the function $(\cdot)^4$ is convex. This leads to: $$\left( \frac{x}{2} + \frac{y}{2} \right)^4 \le \frac12 x^4 + \frac12 y^4$$
Multiply both sides by $16$ and we have: $$(x+y)^4 \le 8x^4 + 8y^4.$$
This process works as long as $(\cdot)^p$ is convex, which holds precisely when $p \ge 1$.
You can show that $(x+y)^p \le x^p + y^p$ when $p < 1$ by other means.
A: Regarding your edit and the question in the comment under OC-Sansoo's answer: (If I understand your issue right, you want reasoning for the choice of vectors?)
Start with the RHS of the inequality we want to show. 
$$  8\left(x^4+y^4\right) = \left(x^4+y^4\right)\left(2^2+2^2\right)$$ 
On the RHS we now have have the vectors $\vec{v}=(x^2,y^2)$ and $\vec{w}=(2,2)$.
Now we apply the CS inequality the first time:
  $$ \left(2x^2+2y^2\right)^2 \leq \left(x^4+y^4\right)\left(2^2+2^2\right)$$
We do the same procedure again with the LHS term in the bracket (the inner product of the vectors $\vec{v}$ and $\vec{w}$):
  $$ 2x^2+2y^2= (1+1)(x^2+y^2)$$
Here we have the vectors $\vec{v}=(1,1)$ and $\vec{w}=(x,y)$.
Applying CS again:
  $$  \left(x+y\right)^2  \leq (1+1)(x^2+y^2)$$
Now we are done, since $(x+y)^4\leq\left(2x^2+2y^2\right)^2$.
On a side note: In your edit the CS inequality should be:
$$|\langle v,w\rangle|^2 \leq \langle v,v \rangle \cdot \langle w,w \rangle$$
A: we have $8(x^4+y^4)-(x+y)^4=7x^4-4x^3y-6x^2y^2-4xy^3+7y^4=(7x^2+10xy+7y^2)(x-y)^2\geq 0$
this is true.
