How can I prove this inequality:$(x+y)^4\le8(x^4+y^4)$? I have to proce that $(x+y)^4\le8(x^4+y^4)$
I have seen that the equality is when $x=y$... 
I have tried to develop $(x+y)^4$ but it leads me nowhere... I think that I have to pass by the intermediate of the averages, but I don't see witchone... Could someone please help?
 A: Hint: If $y=0$ it's obviously true. If $y\ne 0$ then divide both sides by $y^4$ and let $t=\frac{x}{y}$. Then move everything to the RHS and study the function of $t$ which you got. 
A: You can divide both sides by 16, and then use that $f(x)=x^4$ is a convex function (since $f''(x)\ge 0$).
A: your inequality is equivalent to $$8(x^4+y^4)-(x+y)^4=(x-y)^2 \left(7 x^2+10 x y+7 y^2\right)\geq 0 $$
A: $$(x+y)^4= x^4+4x^3y+6x^2y^2+4xy^3+y^4$$
Now, by AM-GM you have
$$
x^3y \leq \frac{x^4+x^4+x^4+y^4}{4} \\
x^2y^2 \leq \frac{x^4+y^4}{2} \\
xy^3\leq \frac{x^4+y^4+y^4+y^4}{4}$$
Thus
$$(x+y)^4 \leq x^4+4 \cdot \frac{x^4+x^4+x^4+y^4}{4} +6 \frac{x^4+y^4}{2} +4 \frac{x^4+y^4+y^4+y^4}{4} +y^4$$
A: My idea: we have equality when $x = y$, so your inequality is true if moving $x$ and $y$ away from equality increases the RHS more than the LHS.
Let $m$ be the midpoint of $x$ and $y$, and $c$ the distance from $m$ to each of them, say $x = m + c$, $y = m - c$, so $c$ is a measure of how different $x$ and $y$ are. We have \[(m + m)^4 = 8(m^4 + m^4),\] as you pointed out, and moreover $(m + c) + (m - c) = m + m$, so the LHS is also $(x + y)^4$, your LHS. So adding $c$ to one term and subtracting it from the other doesn't change the size of the LHS. It only remains to prove that doing the same to the RHS makes it bigger, that is, \[8(m^4 + m^4) \le 8((m + c)^4 + (m - c)^4).\] Expand some brackets and cancel some terms and this should be easy to verify.
