The following question is directly related to this one and involves the resolution (or verification) of an inequality. Again, the first part of my question is dedicated to the creation of the (necessarily physical) context, the second part to the transition to mathematical language, since my goal here is to verify the inequality I'll write at the end of the question.
In Semiclassical's answer, he/she set $y = \dfrac{x}{a+b+c}$ and, after solving the cubic inequality, computed the two solutions as $y_{1}=\dfrac{2}{3}-\dfrac{2}{3}\sin\left[\dfrac{\pi}{6}+\dfrac{\arccos\left(\dfrac{54abc}{(a+b+c)^3}-1\right)}{3}\right]$ and $y_{2}=\dfrac{2}{3}-\dfrac{2}{3}\sin\left[\dfrac{\pi}{6}-\dfrac{\arccos\left(\dfrac{54abc}{(a+b+c)^3}-1\right)}{3}\right]$.
As inferred from the answer, the condition $a, b, c > 0$ such that $\dfrac{abc}{(a+b+c)^3} < \dfrac{1}{27}$ must be satisfied.
In this interval, $y_2 > y_1$, so $x^{\mathrm{max}} = \dfrac{2}{3}(a+b+c) \left\{1-\sin\left[\dfrac{\pi}{6}-\dfrac{\arccos\left(\dfrac{54abc}{(a+b+c)^3}-1\right)}{3}\right]\right\}$ is the point of maximum.
So, the maximum temperature sought is $T_{\mathrm{hotter}}^{\mathrm{max}} = \dfrac{2}{3}(T_1+T_2+T_3) \left\{1-\sin\left[\dfrac{\pi}{6}-\dfrac{\arccos\left(\dfrac{54 \ T_1T_2T_3}{(T_1+T_2+T_3)^3}-1\right)}{3}\right]\right\}$,
where the substitutions$$x^{\mathrm{max}} \mapsto T_{\mathrm{hotter}}^{\mathrm{max}}, \qquad a \mapsto T_1, \qquad b \mapsto T_2, \qquad c \mapsto T_3$$
have been made.
Now, one of the applications of this physical argument is that the maximum solution is reached not through dumping the extracted work into the third system in the form of heat (method 1.), but using a reversible heat pump to move additional heat from the lowest systems (WLoG, with initial temperatures $T_1$ and $T_2$) to cool them even further so that we could obtain an even higher temperature in the hottest system (WLoG, with initial temperature $T_3$) producing no entropy at all (method 2.).
Method 1.
Step 1): Pushing the two systems at temperatures $T_1$ and $T_2$ together.
We start in state $1$ and end in state $2$. The final temperature $T_\mathrm{final}$ is calculated by considering that the net flux of heat is zero (conservation of energy) and the termology fundamental relation:
$$Q_{\mathrm{net}} = 0 \implies \sum_{i = 1}^2 Q_i = 0 \implies Q_1 + Q_2 = 0 \implies C(T_\mathrm{final} - T_1) + C(T_\mathrm{final} - T_2) = 0$$,
from which$$T_\mathrm{final} - T_1 + T_\mathrm{final} - T_2 = 0 \iff \boxed{T_\mathrm{final} = \frac{T_1+T_2}{2}}$$
Step 2): connecting systems at temperatures $T_1$ and $T_2$ by a Carnot engine.
The two systems are connected by a heat engine to extract energy from their temperature difference. This heat engine is reversible,and since the operations of the reversible engine produce no entropy, we have:
$$\Delta S_{\mathrm{net}} = 0 \implies \sum_{i = 1}^2 \Delta S_i = 0 \implies \Delta S_1 + \Delta S_2 = 0 \implies \int \mathrm{d}S_1 + \int \mathrm{d}S_2 = 0 \implies \int\limits_1^{\mathrm{final}} \frac{C\,\mathrm{d}T}{T} + \int\limits_2^{\mathrm{final}} \frac{C\,\mathrm{d}T}{T} = 0,$$
from which
$$C \ln \left(\frac{T_\mathrm{final}}{T_1}\right) + C \ln \left(\frac{T_\mathrm{final}}{T_2}\right) = 0 \implies \ln \left(\frac{T_\mathrm{final}}{T_1}\right) + \ln \left(\frac{T_\mathrm{final}}{T_2}\right) \iff \ln \left(\frac{T^2_\mathrm{final}}{T_1 T_2}\right) = 0,$$
from which
$$\frac{T^2_\mathrm{final}}{T_1 T_2} = 1 \iff \boxed{T_\mathrm{final} = \sqrt{T_1 T_2}}.$$
Even though the reversiblity of the engine produces no entropy, there is entropy generated when the extracted work is converted irreversibly into heat: this extracted work accumulates during the operations of the engine and, after that, the stored energy is expended by heating the two systems equally. So the extracted energy is:
$$Q = C(T_1 - \sqrt{T_1 T_2}) + C(T_2 - \sqrt{T_1 T_2}) = C(T_1 + T_2 - 2\sqrt{T_1 T_2}),$$ so the change in temperature is $\Delta T = T_1 + T_2 - 2\sqrt{T_1 T_2}$. In order to dump heat-engine-extracted work directly into the system at temperature $T_3$ implies we add the change in temperature to $T_3$, having the three bodies the same heat capacity. So:
$$T_{\mathrm{hotter}} = T_1 + T_2 + T_3 - 2\sqrt{T_1 T_2}$$
Method 2. is the one that maximises the hotter body temperature, according to what Callen and another source show graphically.
So we have:
$$T_{\mathrm{hotter}}^{\mathrm{max}} = \dfrac{2}{3}(T_1+T_2+T_3) \left\{1-\sin\left[\dfrac{\pi}{6}-\dfrac{\arccos\left(\dfrac{54 \ T_1T_2T_3}{(T_1+T_2+T_3)^3}-1\right)}{3}\right]\right\}> T_1 + T_2 + T_3 - 2\sqrt{T_1 T_2} = T_{\mathrm{hotter}}, or$$
$$\dfrac{2}{3}\left\{1-\sin\left[\dfrac{\pi}{6}-\dfrac{\arccos\left(\dfrac{54 \ T_1T_2T_3}{(T_1+T_2+T_3)^3}-1\right)}{3}\right]\right\} > 1 - \dfrac{2\sqrt{T_1 T_2}}{T_1+T_2+T_3}$$
Here we can make the step to pure mathematical language. Let
$$T_1 \mapsto a, \quad T_2 \mapsto b, \quad T_3 \mapsto c.$$
We have:$$\boxed{\frac{2}{3}\left\{1-\sin\left[\frac{\pi}{6}-\frac13\arccos\left(\frac{54 \ abc}{(a+b+c)^3}-1\right)\right]\right\} > 1 - \frac{2\sqrt{ab}}{a+b+c}}$$$$ \qquad \forall a,b,c > 0 : \frac{abc}{(a+b+c)^3}< \frac{1}{27}.$$(Can we prove or verify it?)
I had thought of writing the inequality in the form
$$\frac{2}{3}\left\{1-\sin\left[\dfrac{\pi}{6}-\frac13\arccos\left(\dfrac{54 \ abc}{(a+b+c)^3}-1\right)\right]\right\} + \dfrac{2\sqrt{ab}}{a+b+c} - 1 > 0,$$
so that we have to prove something like $g(a_1, a_2, a_3) > 0$. We could pick some function $f$ and fix $f(a_1,a_2)$ and $a_3$, then we can freely change one variable, $a_1$ or $a_2$,since as soon as we change one, we have to change the other in order to make $f(a_1,a_2)$constant, and finally find the minimum of $g(a_1,a_2, a_3)$ when $f(a_1,a_2)$ and $a_3$ are constant so that we can remove a variable.The fact is that the trigonometric nature of the function does not allow the inequality to be simplified trivially, so I am stuck. So I pose the following question:
Are there analytic specific methods to solve this inequality, or at least to verify that it is proven under the given conditions?