<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
$$a^{4/a} + b^{4/b} + c^{4/c} \ge 3.$$

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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] I show the inequality (almost algebraically speaking) for $x\in[0,4/3]$ :
$$4\left(x-1\right)+1\leq x^{\frac{4}{x}}$$
In fact we have a stronger inequality for $x\in[3/4,4/3]$:
$$\left(x^{-\frac{1}{4}}\left(1+\left(\frac{1}{x}+\frac{1}{4}\right)\left(x-1\right)\right)\right)^{4}\leq x^{\frac{4}{x}}\tag{I}$$
To show it $(I)$ we use Bernoulli's inequality simply remarking :
$$x^{\frac{4}{x}}=\left(x^{\frac{1}{x}+0.25-0.25}\right)^4$$
Now see WolframAlpha for a factorization .
Except using derivative I cannot show this sixth degree polynomials is positive for $x\in[3/4,4/3]$
In fact we have clearly for $1\le x\le 4/3$:
$$x^{6}+30x^{5}-687x^{4}+400x^{3}+1632x^{2}-1280x+256>x^{4}+30x^{4}-687x^{4}+400x^{2}+1632x^{2}-1280x+256$$
We can use the second derivative or substitute by $y=x^2$ and factorize again a part.
For $3/4\le x\leq 1$ the sixth degree polynomials is increasing via derivative .
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