<s>[INST] 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 $x, y, z.$
 [/INST] $$\tag{1}\sqrt{2(x+y+z)}\leq a\sqrt{x+y}+b\sqrt{z}$$
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 $b=\sqrt{2}$

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