Show that $\left(\frac{1+\sqrt{5}}{2}\right)^5 > 10$ I have to show that $\left(\frac{1+\sqrt{5}}{2}\right)^5 > 10$.
I've already proven that if $f_n$ is the nth Fibonnaci number, then,
$$f_{n+1} > \left(\frac{1+\sqrt{5}}{2}\right)^{n-1} $$
But I really don't get how to go from this statement to what I need to prove.
Any advices on how to take from here? Any help is welcome.
Thanks.
 A: One option is to expand the power. This is made easier by the fact that $\phi^2 = \phi+1$.
$$
\phi^5 = (\phi+1)(\phi+1)\phi = (3\phi + 2)\phi = 5\phi+3
$$
and since it's easily shown that $\phi > 3/2$, the inequality follows.
If you want to relate to the Fibonacci numbers directly, you can also use $\phi^n = F_n\phi + F_{n-1}$ to get $\phi^5 = 5\phi+3$.
A: Say $x=\sqrt{5}$. Then we have to prove:
$$1+5x+10x^2+10x^3+5x^4+x^5>64x^2 $$
or
$$1+5x+50x+25x> 29x^2$$
or $ 80x> 144$ or $5x>9$ or if we square $125>81$ which is true.
A: Binomially expand
\begin{eqnarray*}
\left( \frac{1+ \sqrt{5}}{2} \right)^5 &=& \frac{1+5 \sqrt{5}  +10 \times 5 +10 \times 5 \sqrt{5}+5 \times 25 +25 \sqrt{5}}{32} \\
&=&\frac{11+5 \sqrt{5}}{2}.
\end{eqnarray*}
Now $125>121 $ square root this and we have the stronger result that 
\begin{eqnarray*}
\left( \frac{1+ \sqrt{5}}{2} \right)^5 > 11. \\
\end{eqnarray*}
A: The light brute force way is sure fire...Just using the elemetary algebra math level: $LHS > RHS \iff \left(1+\sqrt{5}\right)^5 > 2^5\cdot 10\iff(1+\sqrt{5})\left((1+\sqrt{5})^2\right)^2> 2^5\cdot 10\iff(1+\sqrt{5})(6+2\sqrt{5})^2> 2^5\cdot 10\iff (1+\sqrt{5})(56+24\sqrt{5}) > 320\iff 176+80\sqrt{5}> 320\iff \sqrt{5} > \dfrac{144}{80} = 1.8\iff 5 > 1.8^2 = 3.24$ which is clear. If the exponent gets sufficiently big, then the other methods will be much better for sure. 
