TITLE: Algebraic Proof of Geometric Claim
QUESTION [5 upvotes]: I derived this claim geometrically:

If $a<b$ and
$$\sqrt{(x-a)^2+(y-b)^2}<\frac{b-a}{\sqrt2},$$
then $x<y$.

I let $a<b$ and plotted the points $(a,a)$, $(b,b)$, and $(a,b)$. Because they form a right triangle, I figured that the line from $(a,b)$ to the midpoint between $(a,a)$ and $(b,b)$ is the shortest line from $(a,b)$ to the line $f(x)=x$. Therefore, the ball
$$\|(x,y)-(a,b)\|<\frac{b-a}{\sqrt2}$$
is entirely above the line $f(x)=x$, which implies that $x<y$. However, how can I prove the claim algebraically? I am not exceptional with inequalities.

REPLY [5 votes]: Note that 
$$\sqrt2\sqrt{u^2+v^2}\ge |u+v|\ge u+v.$$
So 
$$\frac{b-a}{\sqrt2}>\sqrt{(x-a)^2+(b-y)^2}\ge \frac{b-y+x-a}{\sqrt2}.$$
It then follows that $y>x$.