<s>[INST] when index is irrational number with inequality 
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?
 [/INST] 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.
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