Let $x>0$, show that $$x^{\sqrt{3}}+x^{\frac{\sqrt{3}}{2}}+1\ge 3\left(\dfrac{1+x}{2}\right)^{\sqrt{3}}$$
we consider $$f(x)=2^{\sqrt{3}}(x^{\sqrt{3}}+x^{\dfrac{\sqrt{3}}{2}}+1)- 3(1+x)^{\sqrt{3}}$$ use computer in fact $f(x)\ge 0,x>0$,see Plot we have $$f'(x)=\dfrac{1}{2}\sqrt{3}\left(2^{\sqrt{3}}x^{\dfrac{\sqrt{3}}{2}-1}\left( 2x^{\dfrac{\sqrt{3}}{2}}+1\right)-6(x+1)^{\sqrt{3}-1}\right)$$ for this index irrational numbers,
what approaches do you think, I could take to solving the next step?
On
Let $x = 2z+1$ so that $\frac{1+x}{2} = 1+z$. I also set $p=\sqrt{3}/2$ for commodity. Then the inequality becomes $$ (1+2z)^{2p} + (1+2z)^p + 1 - 3 (1+z)^{2p} \geq 0.$$
The reasoning below this point is false, since $\binom{x}{n}$, for $x$ not an integer and $n > x$, is negative. Actually, even the third coefficient of the power series expansion is $(12 \sqrt{3}-21)/2$, which is negative. I leave this as the trace of a failed attempt...
Expanding as power series of $z$, this becomes $$ \sum \left((2^n-3)\binom{2p}{n}+2^n \binom{p}{n}\right) z^n \geq 0. $$ (Except the constant coefficient, which is zero). So we might show the result at least for $0 \leq z < 1$ (this is the radius of convergence) if we were able to prove that, for $n \geq 1$, $$ (2^n-3) \binom{2p}{n} + 2^n \binom{p}{n} \geq 0, \quad p = \sqrt{3}/2.$$ For $n \geq 2$ this is obvious since $2^n \geq 3$. We manually check that this is also true for $n = 1$: $$ (2^1-3)\binom{2p}{1} + 2^1 \binom{p}{1} = -\sqrt{3}+2 \frac{\sqrt{3}}{2} = 0.$$
This proves the result for $z \in [0, 1[$ and hence $x \in [1,3[$. It remains to show :
the case where $x \geq 3$: this should be easier, since both functions are further apart in this region. (The hardest part was the osculating point at $x=1$).
the case where $0 < x < 1$. However, both sides are quite symmetric with respect to $x \mapsto 1/x$, so that I expect that we may easily deduce it from the case $x > 1$.
first, we find a special point: $\dfrac{1}{3}> \left(\dfrac{1+x}{2}\right)^{\sqrt{3}} \implies x<2\left(\dfrac{1}{3}\right)^{\frac{1}{\sqrt{3}}}-1=0.0606 \implies x<0.06, $ the inequality is always true.
both side take log, we have: $\ln{\dfrac{x^{\sqrt{3}}+x^{\frac{\sqrt{3}}{2}}+1}{3}}\ge \ln{\left(\dfrac{1+x}{2}\right)^{\sqrt{3}}},f(x)=\ln{\dfrac{x^{\sqrt{3}}+x^{\frac{\sqrt{3}}{2}}+1}{3}}-\ln{\left(\dfrac{1+x}{2}\right)^{\sqrt{3}}}$
now we need to prove $f'(x)>0$ when $x>1$ and $f'(x)<0$ when $1>x>0.06$
$f'(x)=\dfrac{\sqrt{3}(2x^{\sqrt{3}}+x^{\frac{\sqrt{3}}{2}}(1-x)-2x)}{(x^{\sqrt{3}}(2x^2+2x)+x^{(\sqrt{3}/2)}(2x^2+2x)+2x^2+2x) }$
$g(x)=2x^{\sqrt{3}}+x^{\frac{\sqrt{3}}{2}}(1-x)-2x$
for $x>1$, $g(x)>0 \iff x^{\frac{\sqrt{3}}{2}}-\dfrac{x}{x^{\frac{\sqrt{3}}{2}}}>\dfrac{x-1}{2} \cap \dfrac{x}{x^{\frac{\sqrt{3}}{2}}}<x^{\frac{\sqrt{3}}{8}}\iff x^{\frac{\sqrt{3}}{2}}-x^{\frac{\sqrt{3}}{8}}>\dfrac{x-1}{2}$
for $0.06<x<1$,prove $g''>0$ which mean $g(x)$ will get max at $g(1)$ or $g(.06)$, if both end $\le 0 \implies g(x) \le 0$
I will not goto details because this is not so difficult.
Edit: a more straight way to prove $g(x)<0 \iff x^{\frac{\sqrt{3}}{2}}-\dfrac{x}{x^{\frac{\sqrt{3}}{2}}}<\dfrac{x-1}{2} $ is:
LHS$=h(x)=x^{\frac{\sqrt{3}}{2}}-\dfrac{x}{x^{\frac{\sqrt{3}}{2}}},h''(x)=x^{-2}(\dfrac{3}{2}-\sqrt{3})(x^{\frac{\sqrt{3}}{2}}-x^{1-\frac{\sqrt{3}}{2}})>0 \cap (x<1) \implies h(x) <y=\dfrac{h(1)-h(0.06)}{1-.06}(x-1)+h(1)=0.562(x-1)<0.5(x-1)$
the tricky here is when $x$ is small enough, $g(x)>0$, that is why we exclude small $x$ at beginning.