how to test the correctness of the following inequality? Can we prove the correctness of the inequality $\sqrt{(x^2+y^2+z^2)}\leq a \sqrt{(x^2+y^2)}+b|z|$ where $x, y, z\in \mathbb{R}.$ What are such $a$ and $b?$
 A: What I am providing is not complete, but rather a way of thinking.
I came a cross this theorem.
Theorem: For any finite dimensional normed space, all norms that you can define are equivalent.
According to the theorem, I would like to define two norms for $\mathbb{R}^3$.
The first one would be the regular one. If $X=(x_1,x_2,x_3)$
$||X||_1=\sqrt{x_{1}^2+x_{2}^2+x_{3}^2}$
The second one is
$||X||_2=\sqrt{x_{1}^2+x_{2}^2}+|x_3|$
To understand the second one, one can prove if there are two normed vector spaces $U$, with norm $||.||_u$, and $V$, with norm $||.||_v$, then the space $U\times V$ is also a normed vector space, where the norm is defined as $||.||_{u\times v}=||.||_u+||.||_v$. This is used to get the second norm above.
Now that we have two norms for $\mathbb{R}^3$ and we know that they are equivalent (according to the first theorem), we use the definition of equivalency. Two norms, in a normed vector space, are equivalent if there exist $0<m<\infty$ and $0<M<\infty$, such that
$m||X||_2\leq ||X||_1\leq M||X||_2$ for all $X\in \mathbb{R}^3$
By plugging in the defined norms, we get
$m(\sqrt{x_{1}^2+x_{2}^2}+|x_3|) \leq \sqrt{x_{1}^2+x_{2}^2+x_{3}^2} \leq M(\sqrt{x_{1}^2+x_{2}^2}+|x_3|)$
Just have a look at the right inequality, which is a bit similar to what you need. But, it only tells you that there is such a finite $M$ that gives you the inequality.
A: If $x=y=0$ we obtain $|z|\leq b|z|$, which gives $b\geq1$.
If $y=z=0$ we obtain $|x|\leq a|x|$, which gives $a\geq1$.
But for $a=b=1$ it's obvious that  $\sqrt{x^2+y^2}+|z|\geq\sqrt{x^2+y^2+z^2}$ is true
for all reals $x$, $y$ and $z$. 
Id est, the best values of $a$ and $b$ are $a=b=1$.
