if $x\in [2009,2010],y\in [2008,2009]$then $(x+y)(\frac{1}{x}+\frac{a}{y})\ge 9,a>0$ find $a_{min}$

Question

Let $x\in [2009,2010],y\in [2008,2009]$ if $$(x+y)(\dfrac{1}{x}+\dfrac{a}{y})\ge 9,a>0$$ Find the $a_{min}$

Answer:$a_{min}=3.5$

$$LHS=(x+y)(1/x+a/y)=1+a+\dfrac{xa}{y}+\dfrac{y}{x}=a+1+\dfrac{1}{t}+ta,t=\dfrac{x}{y}$$however I am unsure

Answer 1

Rearranging, we have $$a \ge y\left(\frac9{x+y}-\frac1x\right)$$ so we have to find the maximum of the above RHS, for $x, y$ in the given domains.

It is not hard, say using calculus for each variable, to show the expression is decreasing in $x$ and increasing in $y$ for the domains involved, so $x=y=2009$ gives the maximum of RHS, leading to $a_{\min} = \frac72$.