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.
 A: 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 $0 \le b \le 1.48$. $\qquad \Box$
A: Not an answer just a remark :
Playing with Vasc's lemma 7.1 see (https://www.isr-publications.com/jnsa/articles-1563-proofs-of-three-open-inequalities-with-power-exponential-functions) we have the inequality for $1\leq x \leq 2$ and $a=x,c=\frac{2}{x}$ then :
$$\left((1-c)^{2}+ac(2-c)-ac(1-c)\ln a\right)^{2}\leq f(x)=x^{\frac{4}{x}}$$
I think it's pretty sharp to show the inequality even if now it seems a bit harder for the eyes .
On the other hand it seems that $f(x)$ is convex on $(0,1]$ so we can use the tangent line method .
I haven't yet an idea to find a lower bound on $(2,3)$.(No need in fact).
Edit we have using a simple bound on logarithm ($1\leq x \leq 2$,$a=x,c=\frac{2}{x}$):
$$\left((1-c)^{2}+ac(2-c)-ac(1-c)\cdot\frac{2\left(a-1\right)}{a+1}\right)^{2}\leq x^{\frac{4}{x}}$$
We can also use the RiverLi's bound or something else like for $x\in (0,1]$  :
$$x^{\frac{4}{x}}\geq g(x)=x^{5}\left(1+\left(\frac{4.25}{x}-5.25\right)\left(x-1\right)\right)$$
And obviously $0<x^{\frac{4}{x}}$.
It works as well .
We can also use Jensen's inequality because it's not hard to show that $g(x)$ is convex on $[1/3,1]$ and we have a polynomial inequality for $1\leq x \leq 2$:
$$2g\left(\frac{3-x}{2}\right)+f\left(x\right)\geq 3$$
We can also show that $f(x)$ is convex on $(0,9/8]$ remarking that :
$$2\cdot1.125^{\frac{4}{1.125}}>3$$
And use Jensen's inequality .
Remains to show the case $a\in(1,1.125),b\in(0,0.875),c\in(1.125,2)$ and for that use Jensen's inequality on the variable $a,b$ and we have with a such constraint :
$$2f\left(\frac{a+b}{2}\right)+f\left(c\right)\geq 3$$
And use the bounds above .
We are done for this sketch of proof .
A: 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 .
A: Using Vasc's paper and @Andreas have already shown that $f\left(x\right)=x^{\frac{4}{x}}$ is convex for $x\in(0,1]$ .So using WLCF-Theorem with an obvious statement we need to show for $1\leq x\leq 2$ :
$$\frac{1}{3}f\left(x\right)+\frac{2}{3}f\left(\frac{3-x}{2}\right)\geq 1$$
In my first answer I show :
we have using a simple bound on logarithm ($1\leq x \leq 2$,$a=x,c=\frac{2}{x}$):
$$g(x)=\left((1-c)^{2}+ac(2-c)-ac(1-c)\cdot\frac{2\left(a-1\right)}{a+1}\right)^{2}\leq x^{\frac{4}{x}}$$
So it's not hard to show for $1\leq x\leq 2$ :
$$\frac{1}{3}g\left(x\right)+\frac{2}{3}j\left(\frac{\left(3-x\right)}{2}\right)\geq 1$$
Where as in my first answer we have for $x\in(0,1]$:
$$j(x)=x^{5}\left(1+\left(\frac{4.25}{x}-5.25\right)\left(x-1\right)\right)\leq f(x)\tag{K}$$
In expanding and using factorization .
To show $(K)$ we have for $x\in[0.8,1]$:
By Bernoulli's inequality :
$$x^{-5+5+\frac{4}{x}}\geq x^{5}\left(1+\left(\frac{4}{x}-5\right)\left(x-1\right)\right)\geq x^{5}\left(1+\left(\frac{4.25}{x}-5.25\right)\left(x-1\right)\right)$$
For the other interval we can use derivative and a bound for logarithm knowing that $j(x)$ can be negative.
It completes this detailed sketch of proof .
Reference :
Cirtoaje, V., Baiesu, A. An extension of Jensen's discrete inequality to half convex functions. J Inequal Appl 2011, 101 (2011). https://doi.org/10.1186/1029-242X-2011-101
