Very basic question about inequalities - show that sign of two expressions are always equal

Question

I want to show that $$\operatorname{sgn} \left( {{h_1}\sqrt {\frac{1}{2}\left( {\frac{1}{{{h_1}}} + \frac{1}{{{h_2}}}} \right)} - \sqrt {{h_2}} } \right) = \operatorname{sgn} \left( {{h_1} - {h_2}} \right)$$

when $h_1, h_2>0$.

Let $y={{h_1}\sqrt {\frac{1}{2}\left( {\frac{1}{{{h_1}}} + \frac{1}{{{h_2}}}} \right)} - \sqrt {{h_2}} }$.

Using the definition $\operatorname{sgn} x = \left\{ \begin{gathered} 1,\;\;\,\;\;x > 0 \hfill \\ 0,\;\;\;\;x = 0 \hfill \\ - 1,\,\;\;x < 0 \hfill \\ \end{gathered} \right.$,

then I need to show that

$$\operatorname{sgn} y = \left\{ \begin{gathered} 1,\;\;y > 0 \hfill \\ 0,\;\;y = 0 \hfill \\ - 1,\;\;y < 0 \hfill \\ \end{gathered} \right\} = \operatorname{sgn} \left( {{h_1} - {h_2}} \right) = \left\{ \begin{gathered} 1,\;\;{h_1} > {h_2} \hfill \\ 0,\;\;{h_1} = {h_2} \hfill \\ - 1,\;\;{h_1} < {h_2} \hfill \\ \end{gathered} \right\}.$$

I have shown that if $h_1>h_2$ then $y>0$ as well, since ${h_1}\sqrt {\frac{1}{2}\left( {\frac{1}{{{h_1}}} + \frac{1}{{{h_2}}}} \right)} - \sqrt {{h_2}} > \sqrt {{h_1}} - \sqrt {{h_2}} > 0$ if $h_1>h_2$.

My question is this: is it also necessary to show that if $y>0$ then $h_1>h_2$ in order to conclude that $y>0 \Leftrightarrow h_1>h_2$ for the first row of the equation above?

(I have shown that this is true as well, since if $y>0$ then ${h_1}{h_2} + h_1^2 > 2h_2^2$, completing the square gives ${\left( {{h_1} + \frac{{{h_2}}}{2}} \right)^2} > \frac{9}{4}h_2^2$ which implies $h_1>h_2$)

Answer 1

Your question is this: is it also necessary to show that if $y>0$ then $ h_1>h_2$ in order to conclude that $ y>0\iff h_1>h_2$ for the first row of the equation above?

The answer is yes.

You have to show both sides to make sure you have if and only if for the first row of the equation.

Answer 2

Multiplying by the conjugate, we see that $$ h_1\sqrt{\frac{1}{2}\left(\frac{1}{h_1} + \frac{1}{h_2}\right)}-\sqrt{h_2} $$ has the same sign as $$ h_1^2\frac{1}{2}\left(\frac{1}{h_1} + \frac{1}{h_2}\right)-h_2= \frac{h_1h_2+h_1^2-2h_2^2}{2h_2}=\frac{(h_1-h_2)(h_1+2h_2)}{2h_2} $$