Given positive variable $(x,y,a,b)$ where $x+y=1$, how does one "slickly" prove the following inequality? $$f(x,y) := \frac{xa+yb}{\sqrt{xa^2+yb^2}}\ge \frac{2\sqrt{ab}}{a+b}.$$ or simply $$f(x,y) := \frac{xa+yb}{\sqrt{xa^2+yb^2}}\ge 2\sqrt{ab},$$ with constraint $a+b=1$.
I can prove this by differentiating $f(x,1-x)$ with respect to $x$ and show it first decrease then increase on $[0,1]$, or multiply both sides by the denominators then square them and examine the resulting quadratic equation. However, there is symmetry with this problem where the minimum of $f(x,y)$ is achieved at $xa=yb$. So I am seeking a more direct and "beautiful" approach to exploit this symmetry or use something like convexity, Cauchy-Schwarz inequality, etc, the methods I can use to generalize this problem.
Further, without knowing the minimum of $f(x,1-x)$ given $(a,b)$, how would one with a slick method and without the aforementioned differentiation, find the minimum?
In homogeneous form, the inequality can be written as $$(xa+yb)^2(a+b)^2 \ge 4ab(x+y)(xa^2+yb^2)$$
which is the same as $(a-b)^2(ax-by)^2 \ge 0$, from which the inequality and equality conditions are both obvious.