$x^{4}+ y^{4}+ z^{4}\leq 2(\left | xy \right |^{2}+ \left | yz \right |^{2}+ \left | zx \right |^{2})$

Question

Given real numbers $x, \,y,\,z$ such that:

$$x^{4}+ y^{4}+ z^{4}\leq 2(\left | xy \right |^{2}+ \left | yz \right |^{2}+ \left | zx \right |^{2})$$

I want to prove:

$$2(\left | x \right |+ \left | y \right |+ \left | z \right |)= \sqrt{(x+ y-z)^{2}}+ \sqrt{(y+ z- x)^{2}}+ \sqrt{(z+ x-y)^{2}}+ \sqrt{(x+ y+ z)^{2}}$$

I found some equalities:

We have:

$$(x+ y-z)^{2}+ (y+ z- x)^{2}+ (z+ x-y)^{2}+ (x+ y+ z)^{2}= 4(x^{2}+ y^{2}+ z^{2})$$

$$4(x^{2}+ y^{2}+ z^{2})^{2}= 2[(x^{4}+ y^{4}+ z^{4})- (x^{2}y^{2}+ y^{2}z^{2}+ z^{2}x^{2})]\leq 2(x^{2}y^{2}+ y^{2}z^{2}+ z^{2}x^{2})$$

Sorry, but I can't continue. Help me! Thanks!

Answer 1

If we'll replace $x$ on $-x$ then the condition does not change and the needed does not change.

We can do it with any variables, which says that we can assume that all variables are non-negatives.

In another hand, the condition and the needed they are symmetric.

Thus, it's enough to solve our problem for $x\geq y\geq z\geq0.$

Now, $$0\leq\sum_{cyc}(2x^2y^2-x^4)=(x+y+z)\prod_{cyc}(x+y-z),$$ which gives $y+z\geq x$.

Hence, we need to prove that $$2(x+y+z)=\sum_{cyc}|x+y-z|+|x+y+z|$$ or $$2(x+y+z)=\sum_{cyc}(x+y-z)+x+y+z,$$ which is obvious.