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I'm studying physics (sorry) and can't for the life of me express the following as greater than one. All $T$s are values for temperature, and thus greater than 0.

$\frac{T_0^2}{T_HT_C}$ (should be) > 1 where $T_0=\frac{1}{2}(T_H+T_C)$.

Very appreciative of any solutions or suggestions.

For context, taken from the 'hot bricks' problem in thermodynamics.

N Godrich
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3 Answers3

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From a.m.-g.m. inequality, we obtain $${\rm arithmetic}\,{\rm mean}\left ( T_{H}, T_{C} \right )\geq{\rm geometric}\,{\rm mean}\left ( T_{H}, T_{C} \right )$$ That is $$\frac{T_{H}+ T_{C}}{2}\geq\sqrt{T_{H}T_{C}}\Leftrightarrow\frac{T_{0}}{\sqrt{T_{H}T_{C}}}> 1$$

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$\frac{T_0^2}{T_HT_C}>1$ if and only if $T_0^2>T_HT_C$ if and only if $T_0^2-T_HT_C>0$. Since $$T_0^2-T_HT_C = \frac14T_H^2 - \frac12 T_H T_C + \frac14T_C^2 = \frac14(T_H-T_C)^2 \geq0,$$ this is true if $T_H \neq T_c$.

Steven
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Notice that for some $x$, we have $T_{H} = T_{0} +x$ and $T_{C} = T_{0} - x$. However then $T_{H}T_{C} = T_{0}^{2} - x^{2}$.
Now do the division.

Edit follows

Since the $T$'s are all $> 0$, we have $T_{0}^{2} \geq T_{0}^{2} - x^{2} > 0$. But then $\frac{T_{0}^{2}}{T_{0}^{2} - x^{2}} \geq 1$.

Jay
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