Áp dụng BĐT Cauchy ta có:
$2^{2\sin x}+2^{\tan x}\ge2\sqrt{2^{2\sin x}.2^{\tan x}}=2^{\frac{2\sin x+\tan x}{2}+1}$
Ta chỉ cần chứng minh: $2^{\frac{2\sin x+\tan x}{2}+1}>2^{\frac{3x}{2}+1} \Leftrightarrow 2\sin x+\tan x>3x$
Xét hàm số: $f(x)=2\sin x+\tan x-3x,x\in\left(0;\,\dfrac{\pi}{2}\right)$
Ta có:
$f'(x)=2\cos x+\dfrac{1}{\cos^2x}-3\ge3\sqrt[3]{\cos x.\cos x.\dfrac{1}{\cos^2x}}-3=0$
hay $f'(x)\ge0,\forall x\in\left(0;\,\dfrac{\pi}{2}\right) \Rightarrow f(x)$ đồng biến trên $\left(0;\,\dfrac{\pi}{2}\right)$
Suy ra: $f(x)>f(0)=0 \Leftrightarrow 2\sin x+\tan x>3x,\forall x\in\left(0;\,\dfrac{\pi}{2}\right)$