How to validate the following inequality?

Question

$\sqrt{2(x+y+z)}\leq a\sqrt{x+y}+b\sqrt{z}.$ I want to find the least values of $a$ and $b$ for which the above inequality holds good for all nonnegative real values of $x, y, z.$

Answer 1

$$\tag{1}\sqrt{2(x+y+z)}\leq a\sqrt{x+y}+b\sqrt{z}$$

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 $b=\sqrt{2}$

Answer 2

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.

Answer 3

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.