Find the maximum $k$ such that $x^2(3-2x)(2x^k+(3-2x)^k)\le 3$ for all $0\le x\le 1$

Question

Find $k_{\max}$,such $$0\le x^2(3-2x)(2x^k+(3-2x)^k)\le 3$$

which satisfies the ineqlities for all $x\in [0,1]$

A sketch of my thoughts

since $$x^2(3-2x)>0\Longrightarrow 2x^k+(3-2x)^k\ge 0$$ it is clear for $k\in R$ and other case it's not easy to solve

Answer 1

This answer is incomplete.

Put $$f(x,k)=x^2(3-2x)(2x^k+(3-2x)^k).$$ Since $$3-f(x,2)=3\left(x-1\right)^2\left(4x\left(x-\frac 58\right)^2+\frac {7x}{16}+(1-x^2)\right)\ge 0,$$ $$k_{max}\ge 2.$$

But a computational evidence suggests that $$k_{max}=2.25806...,$$ because of the graphs of $f(x,2.25806)$ below and an inequality $$f(0.85,2.25807)\simeq 3.00000027895282>3.$$