The inequality $$y^2 - xy + x + 2y + 5 \geq k\,(\sqrt{3x} + \sqrt{y})$$ holds for all $x$ and $y$ such that $0\leq x,y\leq 3$. Find maximum $k$.
The answer is $2$. Can someone please solve it? Thanks in advance.
My thoughts on this problem: Since $0\leq x,y\leq 3$, I try to use the fact that $(3-x)(3-y)\geq 0$ and $(3-x)(y-3)\leq 0$.
My second thought was: If this $$y^2 - xy + x + 2y + 5 \geq k\,(\sqrt{3x} + \sqrt{y})$$ is true, then $$x^2 - xy + y + 2x + 5 \geq k\,(\sqrt{3y} + \sqrt{x})$$ is also true, and I sum them together, but couldn't get anywhere from there too.
Hint: Since this inequality must be true for all $0\leq x,y\leq 3$ it must be true also for some specific values. Putting $x=3$ and $y=1$ we get $$8\geq 4k\implies k\leq2$$
so we must only prove $k=2$ works...
Further hint: We have $$2\sqrt{3x}+2\sqrt{y}\leq (x+3)+(y+1)$$
Final hint: Is it true (for $x,y\in [1,3]$):
$$y+1+{1\over y}\geq x ?$$