Show properties in $\mathbb{R}^n$ I want to show the following properties in $\mathbb{R}^n$.


*

*$2 \|x\|^2+2 \|y\|^2=\|x+y\|^2+\|x-y\|^2$

*$\|x-y\| \|x+y\|\leq \|x\|^2+ \|y\|^2$


I have tried the following:


*

*$$\|x+y\|^2+\|x-y\|^2=(x+y) \cdot (x+y)+(x-y) \cdot (x-y)=\|x\|^2$$
$$+ x \cdot y+ y \cdot x+ \|y\|^2+ \|x\|^2-x \cdot y-y \cdot x+ \|y\|^2=2 \|x\|^2+ 2 \|y\|^2$$

*$\|x-y\| \|x+y\| \leq (\|x\|+\|y\|)^2=\|x\|^2+ 2 \|x\| \|y\|+\|y\|^2$
But the latter isn't $\leq \|x\|^2+ \|y\|^2$. Have I done something wrong?
 A: The second inequality is true iff the squared inequality is true, i.e iff
$$\|x-y\|^2\|x+y\|^2\leq (\|x\|^2+\|y\|^2)^2$$
You have
$$\|x-y\|^2\|x+y\|^2=(\|x\|^2-2(x,y)+\|y\|^2)(\|x\|^2+2(x,y)+\|y\|^2)=(\|x\|^2+\|y\|^2)^2-4(x,y)^2\leq (\|x\|^2+\|y\|^2)^2$$
Now if you are wandering why your inequality is correct, you can take square root of both sides of this inequality, because they are positive, and because the square root is monotonically increasing function.
A: $$(\|x-y\|\|x+y\|)^2=(x-y)^2\cdot(x+y)^2$$ $$=(x^2-2xy+y^2)\cdot(x^2+2xy+y^2)$$
$$=x^4-2x^2y^2+y^4$$ I skipped a test in trivial calculation so show this in your workings. Now as $x^2$ and $y^2$ are always positive we find: $$(\|x-y\|\|x+y\|)^2\leq x^4+y^4=\|x\|^4+\|x\|^4$$ As both sides are positive and $\sqrt{t}$ is monotone increasing on that interval then $$\|x-y\|\|x+y\|\leq \sqrt{\|x\|^4+\|x\|^4}\leq \sqrt{\|x\|^4}+\sqrt{\|y\|^4}$$
Then you are done. The last inequality follows from an inequality involving the sum of square roots. It's simple enough to prove so I'll leave that to you if you have not encountered it before.
EDIT:
Notice that 
$$(\|x-y\|\|x+y\|)^2=(x-y)^2\cdot(x+y)^2=\big((x-y)(x+y)\big)^2$$ 
$$=(x^2-y^2)^2=(\|x\|^2-\|y\|^2)^2\leq (\|x\|^2+\|y\|^2)^2$$
which leads directly to the result using the logic as shown above.
A: The first is ok. The second is flawed (at least, I don't see the inequality you use there). Square the inequality and prove this.
EDIT: The second "proof" isn't flawed. The OP just used the triangle inequality here two times (see the comments below).
