Proving that $8^x+4^x\geq 5^x+6^x$ for $x\geq 0$. I want to prove that $$8^x+4^x\geq 6^x+5^x$$ for all $x\geq 0$. How can I do this?
My attempt:
I try by AM-GM: $$8^x+4^x\geq 2\sqrt{8^x4^x}=2(\sqrt{32})^x.$$
However, $\sqrt{32}\approx 5.5$ so I am not sure if $$2(\sqrt{32})^x\geq 5^x+6^x$$ is true.
Also, I try to compute derivatives but this doesn't simplify the problem. What can I do?

Note that $8^x=(6^x)^{\log_6(8)}$ so that \begin{equation}\tag 1\label 14^x+8^x-6^x-5^x\geq 8^x-6^x-5^x\to\infty\end{equation} as $x\to\infty$. 
I will show that the only non-negative solution to \begin{equation}\tag 2\label 28^x+4^x=6^x+5^x\end{equation} is $x=0$. Note that $$8^x+4^x=6^x+5^x\iff 8^x-6^x=5^x-4^x.$$
By the Mean value Theorem, we have $8^x-6^x=2x\cdot c^{x-1}$ and $5^x-4^x=x\cdot d^{x-1}$ for some $c\in[6,8]$ and $d\in[4,5]$.
So we have that \eqref{1} is equivalent (for suitable $c$ and $d$) to $$2x\cdot c^{x-1}=x\cdot d^{x-1}.$$ For $x>0$ this is equivalent to $$2=\left(\frac{d}{c}\right)^{x-1}$$ which is impossible since $d\le5<6\le c$. Hence, $x=0$ is the only non-negative solution to \eqref{1}. 
From the intermediate value Theorem (and from \eqref{2}), it follows that $8^x+4^x-6^x-5^x>0$ for all $x>0$.
