We claim that $y>z>x$ Lemma 1: $y>z$ Proof: Let $f(x)=\frac{1}{x}-\frac{1}{\sqrt{x}}$, note that $f^\prime (x)=\frac{\sqrt{x}-2}{2x^2}<0, \forall x\in(0, 4)$ therefore $f(x)$ is decreasing in that interval, as $1>b>b-a>0$ then $f(b-a)=\frac{1}{b-a}-\frac{1}{\sqrt{b-a}}>\frac{1}{b}-\frac{1}{\sqrt{b}}=f(b)$ $\hspace{20cm}\blacksquare$ Lemma 1: $z>x$ Proof: As $g(x)=\frac{1}{\sqrt{x}}$ is convex for $x>0$, by Jensen $$\frac{g(b-a)+g(b+a)}{2}>g\left(\frac{(b-a)+(b+a)}{2}\right) \iff$$ $$\frac{\frac{1}{\sqrt{b-a}}+\frac{1}{\sqrt{b+a}}}{2}>\frac{1}{\sqrt{\frac{(b-a)+(b+a)}{2}}}=\frac{1}{\sqrt{b}}$$ $\hspace{20cm}\blacksquare$

$y>z$ because $y=z(\frac{1}{\sqrt{b-a}}+\frac{1}{\sqrt{b}})>z$ $z>x$ because $z^2 =\frac{1}{b-a}+\frac{1}{b+a}+\frac{2}{\sqrt{b^2-a^2}}=\frac{2b}{b^2-a^2}+\frac{2}{\sqrt{b^2-a^2}}>\frac{4}{b}=x^2$

Nice.

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