Prooving an inequality in thermodynamics

Question

While solving an exercise in thermodynamics, trying to prove the entropy increased for some configuration, I reduced my problem to the following:

$$ \Delta S = T_{final}^{1/2} - \frac{T_1^{1/2} + T_2^{1/2}}{2}, T_{final} = \Big( \frac{T_1^{3/2} + T_2^{3/2}}{2} \Big)^{2/3} $$ Where $T_1 , T_2$ are real positive numbers such that $T_1 > T_2 > 0$, and I wish to prove that $\Delta S$ is greater than $0$.

I have tried using $L_p$ norms and inequalities such as Hölder and Minkowski inequality or triangle inequality but nothing worked for me. I tested this inequality numerically and as far as I can tell it holds up.

Answer 1

For $p\not=0$, let $M_p(x,y)=(x^p+y^p)^{1/p}$ with $x,y>0$ then, by the power mean inequality, if $p\geq q$ then $$M_{p}(x,y)\geq M_{q}(x,y).$$ Equality holds if and only if $x=y$.

Therefore, for $T_1 > T_2 > 0$, $$\Delta S=\Big( \frac{T_1^{3/2} + T_2^{3/2}}{2} \Big)^{1/3} - \frac{T_1^{1/2} + T_2^{1/2}}{2}=M_{3/2}(T_1,T_2)^{1/2}-M_{1/2}(T_1,T_2)^{1/2}>0.$$

Answer 2

Consider $$\Delta S = \Bigg( \frac{T_1^{3/2} + T_2^{3/2}}{2} \Bigg)^{1/3} -\Bigg( \frac{T_1^{1/2} + T_2^{1/2}}{2}\Bigg)$$

Let $T_2=k^2 \,T_1$ with $(0 < k < 1)$ which gives $$ \frac 2 {\sqrt{T_1}} \Delta S = \sqrt[3]{4(k^3+1)}-(k+1)$$ and the rhs is always positive (decreasing from $[2^{2/3}-1]$ to $0$) and its derivative does not cancel in the range.