It is very well-known that given positive real numbers $a,b$, then $$ \dfrac{a+b}{2} \le \sqrt{\dfrac{a^2+b^2}{2}} , $$ and equality holds if and only if $a=b$.
Is there an analogous "weighted" inequality? That is, if $0<\lambda<1$ and $\mu=1-\lambda$, can we say that for every positive real numbers $a,b$, $$ \lambda\cdot a + \mu\cdot b \le \sqrt{\lambda\cdot a^2 + \mu\cdot b^2} , $$ and that equality holds if and only if $a=b$? If so, does this result have a name?