points on the curve $x^2+2y^2=6$ whose distance from the line $x+y-7=0$ is minimum 
Find point on the curve $x^2+2y^2=6$ whose distance from the line $x+y-7=0$ is, minimum

$\bf{My\; Try::}$ Let $(x,y)$ be any point on the curve $x^2+2y^2=6\;,$ Then we have to
minimize $\displaystyle \left|\frac{x+y-7}{\sqrt{2}}\right|$
Using $\bf{Cauchy\; Schwarz}$ Inequality
$$\left[x^2+\left(\sqrt{2}y\right)^2\right]\cdot \left[1^2+\left(\frac{1}{\sqrt{2}}\right)^2\right]\geq (x+y)^2$$
So $$6\times \frac{3}{2}\geq (x+y)^2\Rightarrow (x+y)^2\leq 3^2\Rightarrow-3 \leq  (x+y)\leq 3$$
So We get $$\frac{-3-7}{\sqrt{2}}\leq \frac{x+y-7}{\sqrt{2}}\leq \frac{3-7}{\sqrt{2}}\Rightarrow -5\sqrt{2}\leq \frac{x+y-7}{\sqrt{2}}\leq -2\sqrt{2}$$ 
and equality hold when $\displaystyle \frac{x}{1} = \frac{2y}{1}$
Now i did not understand how can i calculate $\displaystyle \min \left|\frac{x+y-7}{\sqrt{2}}\right|,$ Help me thanks 

Let $t = x+y$, you've shown $-3 \le t \le 3$, thus the function $f(t) =\dfrac{ (t-7)^2}{2}$ has a graph a parabola with the axis of symmetry at $x = 7$, and it is decreasing on $[-3,3]$, thus the min is $f(3) = 8$. Thus the min distance is $2\sqrt{2}$. 
