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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.

wjmolina
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1 Answers1

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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 $y>x$.

Quang Hoang
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