inequality with exponentials!

Question

Proove this inequality: $2(\sqrt{3}+1)^{-x}+2^x(2+\sqrt{3})^x>3$ I tried using Bernoulli, but it doesn't work. Nor does the inequality of arithmetic and geometric means. Please help!

Answer 1

HINT: take $a=(\sqrt3 +1)^x$
$\to a^2=(3+1+2\sqrt3)^{2x}=(2(2+\sqrt3))^{x}$ $$2(\sqrt{3}+1)^{-x}+2^x(2+\sqrt{3})^x>3$$ so rewrite as $$2a^{-1}+(a^2)>3$$and obviously $a>0$

now : note that $$2a^{-1}+(a^2)>3 \space \space\space\space \times a\\ 2+a^3 >3a \to a^3-3a+2 >0\\ (a-1)(a^2+a-2)>0\\(a-1)(a-1)(a+2)>0\\(a-1)^2(a+2)>0 $$ to end of the proof ,we need to say $(a-1)^2 \geq 0$and $a>0 \to a+2>2>0$