<s>[INST] Prove that $a^{4/a} + b^{4/b} + c^{4/c} \ge 3$ 
Let $a, b, c > 0$ with $a + b + c = 3$. Prove that
$$a^{4/a} + b^{4/b} + c^{4/c} \ge 3.$$

This question was posted recently, closed and then deleted, due to missing of contexts etc.
By https://approach0.xyz/, the problem was proposed by Grotex@AoPS.
My strategy is to split into many cases.
WLOG, assume that $a \ge b \ge c$.
If $a \ge 8/5$, true.
If $a \le 10/7$, let
$f(x) = x^{4/x} - 1  - 4(x - 1)$.
We have $f(x) \ge 0$ for all $x \in (0, 10/7)$.
If $10/7 < a < 8/5$ and $b \ge 4/5$, true.
(I stopped here since this approach is ugly. Actually, the proof of $x^{4/x} - 1  - 4(x - 1) \ge 0$ for all $x\in (0, 10/7)$ is complicated.)
I hope to see nice proofs.
 [/INST] Before the actual proof, let's start with a discussion of $f(x) = x^{4/x}$ which is needed later.
We have
$ df(x) /dx = 4 x^{4/x - 2} (1 - \log(x))$ so $f(x)$ is rising for all $0 < x  < e$ which is all we need to consider below. Further,
$$ d^2f(x) /dx^2 = 4 x^{4/x - 4} (-3 x + 4 \log^2(x) + 2 (x - 4) \log(x) + 4)$$
To interpret this result for positive and negative ranges, consider $ h(x) =  -3 x + 4 \log^2(x) + 2 (x - 4) \log(x) + 4$. Clearly, $h(x) {{(x\to 0) } \atop{\longrightarrow}} \infty $. Further,
$$ h'(x) = (-x + 2 (x + 4) \log(x) - 8)/x
$$
which remains negative for at least $0 < x < 2$. So if we have a value where $d^2f(x) /dx^2$ changes sign in the range $0 < x < 2$, this will be the only sign change there. The sign change indeed occurs near $x= 1.125$.
We can now describe the behavior of $f(x) = x^{4/x}$ for $0 < x  < 2$. It is rising, it is convex for   $0 < x  < x_0 $ and concave for $ x_0 < x  < 2$ where $x_0≈1.125$ is the unique real solution of $h(x)=0$ on $(0,2)$.
Now consider $g(x) = -3 + 4x$. Note $g(x)$ touches $f(x) = x^{4/x}$ at $x=1$, since $g(1) =f(1) = 1$ and $g'(1) = f'(1) = 4$. Since $x^{4/x}$ is convex at $x=1$ and turns concave at $x \simeq 1.125$, we have an "S-shape" and hence,  there is at most one intersection $g(x) = f(x)$ in the range  $1.125 < x  < 2$. By inspection we   have that this occurs near $\bar{x} = 1.429$ . So $x^{4/x} \ge g(x)$ for $x < \bar{x} = 1.429$ and hence, for $a,b,c < \bar{x}$, $a^{4/a} + b^{4/b} + c^{4/c} \ge  -9 + 4(a+b+c) =3$.
As OP already noted, only $1.52 > a \ge b \ge c$ must then be considered (since $1.52^{4/1.52} > 3$) . Now if $1.52 > a > \bar{x} = 1.429$, then $a^{4/a} \ge \bar{x}^{4/\bar{x}} \simeq 2.71$ and $b +c \ge 3 - 1.52 = 1.48$. Hence it remains to be shown that for  $b +c = 1.48$, that $b^{4/b} + c^{4/c} \ge 3 - 2.71 = 0.29$. This can be established by direct calculation, since indeed $b^{4/b} + (1.48-b)^{4/(1.48-b)} > 0.29$ for all $0 \le b \le 1.48$. $\qquad \Box$
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