Inequality $\sqrt{x^2+y^2+z^2+2yz}\leq a \sqrt{x^2+y^2}+b|z|,$

Question

How to prove this inequality: $\sqrt{x^2+y^2+z^2+2yz}\leq a \sqrt{x^2+y^2}+b|z|,$ for all real numbers $x, y, z$ for some non negative values of $a$ and $b.$ What are the minimum values of $a$ and $b.$

Answer 1

Let $z=0$. Hence, $a\geq1$.

Let $x=y=0$. Hence, $b\geq1$.

We'll prove that $a=b=1$ are valid.

Indeed, we need to prove that $$\sqrt{x^2+y^2+z^2+2yz}\leq\sqrt{x^2+y^2}+|z|$$ or after squaring of the both sides $$x^2+y^2+z^2+2yz\leq x^2+y^2+2|z|\sqrt{x^2+y^2}+z^2$$ or $$\sqrt{z^2(x^2+y^2)}\geq yz,$$ which is true because $$\sqrt{z^2(x^2+y^2)}\geq\sqrt{y^2z^2}=|yz|\geq yz.$$ Done!

Answer 2

We have $2yz\le y^2+z^2$, hence :

$$x^2+y^2+z^2+2yz\le x^2+2y^2+2z^2\le2(x^2+y^2)+2z^2$$

and finally :

$$\sqrt{x^2+y^2+z^2+2yz}\le\sqrt2\sqrt{x^2+y^2}+\sqrt2\vert z\vert$$

because, for every $a,b\ge 0$ :

$$\sqrt{a+b}\le\sqrt a+\sqrt b$$

This doesn't prove the minimality ...