How to validate the following inequality? $\sqrt{2(x+y+z)}\leq a\sqrt{x+y}+b\sqrt{z}.$ I want to find the least values of $a$ and $b$ for which the above inequality holds good for all nonnegative real values of $x, y, z.$
 A: For$x=y=0$ we get $b\geq\sqrt2$.
For $x=z=0$ we get $a\geq\sqrt2$ and 
$$\sqrt{2(x+y+z)}\leq\sqrt{2(x+y)}+\sqrt{2z}$$ 
is obvious after squaring of the both sides.
A: Assume that $z \neq 0$ and $x, y$ are not simultaneously zero. (For example, if $z$ is zero, $b$ can be any arbitrary negative number.)
Consider the equation $$b = -\sqrt{\frac{x + y}{z}}a + \sqrt{\frac{2(x + y + z)}{z}}.$$
Let $a^* = \sqrt{\frac{2(x + y + z)}{x + y}}$ and $b^* = \sqrt{\frac{2(x + y + z)}{z}}$, i.e., $a^*$ is the $a$-intercept and $b^*$ is the $b$-intercept. The least value of $a$, $b$ (hence $a + b$) is then occurring at $(a^*, 0)$ if $a^* < b^*$ and $(0, b^*)$ otherwise.
A: $$\tag{1}\sqrt{2(x+y+z)}\leq a\sqrt{x+y}+b\sqrt{z}$$
Let us proceed by successive equivalent propositions, reaching at the end two unique values for $a$ and $b$.
First of all, you do not need three variables: you can amalgamate $x$ and $y$ into a single variable $t:=x+y$.
$$\tag{2}\sqrt{2(t+z)}\leq a\sqrt{t}+b\sqrt{z}$$
Set now $T=\sqrt{t}$ and $Z=\sqrt{z}.$
$$\tag{3}\sqrt{2(T^2+Z^2)}\leq aT+bZ.$$
As all the quantities in (3) are positive, we have the following equivalent inequality:
$$\tag{4}2(T^2+Z^2)\leq a^2T^2+b^2Z^2+2abTZ.$$
$$\tag{5} (a^2-2)T^2+(b^2-2)Z^2+2abTZ \geq 0.  \ \ \text{for any positive} \  T,Z $$
Dividing by $Z^2$, and setting $V=T/Z$, one gets the quadratic inequality:
$$\tag{6}(a^2-2)V^2+(2ab)V+(b^2-2) \geq 0 \  \ \text{for any positive} \  V$$
Taking into account


*

*the sign of discriminant $\Delta=4a^2b^2-4(a^2-2)(b^2-2)=8(a^2+b^2-2)$ and

*the sign of dominant coefficient $a^2-2,$
the only possible case is when $a^2-2\geq0$ and $a^2+b^2-2\leq 0$.
The only (positive) values of $a$ and $b$ fulfilling these two conditions are

$a=\sqrt{2}$ and $b=\sqrt{2}$

