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?
Hint. Let $f(t)=t^x$ then by the Mean Value Theorem there is $t_1\in (6,8)$ such that
$$f(8)-f(6)=f'(t_1)(8-6)\Leftrightarrow 8^x-6^x=2xt_1^{x-1}.$$
Similarly there is $t_2\in (4,5)$ such that
$$f(5)-f(4)=f'(t_2)(5-4)\Leftrightarrow 5^x-4^x=xt_2^{x-1}.$$
It remains to show that for $x\geq 0$
$$2xt_1^{x-1}\geq xt_2^{x-1}.$$
The best and the easiest way to prove it is using Induction . 

 Base Case :  

For $x=1$ , $ 8 + 4 \gt 6+5$

Induction Step :

Let us assume that $8^k +4^k \ge5^k + 6^k$ . This implies that $$8^k \ge (5^k + 6^k) - 4^k$$

Multiplying both the sides by $8$ , we get

$$\color{#f14}{8^{k+1} \ge 8\cdot5^k +8\cdot6^k -8\cdot4^k} \quad \quad \text{(1.)}$$

Now all that is left is to prove that the R.H.S is bigger than $5^{k+1}+6^{k+1}-4^{k+1}$.We prove it by assuming it is true . Then ,

$$\begin{align}8\cdot5^k +8\cdot6^k -8\cdot4^k &amp; \ge 5^{k+1}+6^{k+1}-4^{k+1} \\
5^k(8-5) + 6^k(8-6) +4^k(4-8) &amp; \ge 0  \\ 
3\cdot 5^k +2\cdot6^k - 4\cdot4^k &amp; \ge 0 \\ 
\color{#2c0}{\left(\frac 34\right)\cdot \left(\frac 54\right)^k + \left(\frac 24\right)\cdot \left(\frac 64\right)^k }&amp; \color{#2c0}{\ge  0 }\quad \quad \text{(2.)}\end{align}$$

which is obviously true for $k \ge 0$ . Hence our initial assumption was true.

Now combining $(1.)$ and $(2.)$ , we get ,

$$\color{#f14}{8^{k+1}} \ge \color{navy}{8\cdot5^k +8\cdot6^k -8\cdot4^k} \ge \color{#2c0}{5^{k+1}+6^{k+1}-4^{k+1}} $$

or $$ 8^{k+1}  \ge 5^{k+1} + 6^{k+1} -4^{k+1} \implies \boxed { 8^{k+1} + 4^{k+1} \ge 5^{k+1} + 6^{k+1}}$$

Which completes our induction.