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!
inequality with exponentials!
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$\begingroup$
inequality
exponential-function
1 Answers
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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 $a>0 \to a+2>2>0$
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0Are you sure that your expression for $a^2$ is correct? It seems that you doubled the exponent *and* squared the term inside the parentheses. – 2017-02-22
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0@MartinR ,I'll correct it , thank you – 2017-02-22