Proving that $8^x+4^x\geq 5^x+6^x$ for $x\geq 0$.

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 $x\geq 0$ $$2xt_1^{x-1}\geq xt_2^{x-1}.$$

Use: $$ 4^x \left(\left(\frac 32\right)^x-1\right) \left(\left(\frac 43\right)^x-1\right) \ge 0 $$ for $x\ge 0$.

The function $u\mapsto u+{1\over u}$ is increasing for $u\geq1$. Therefore we have for all $x\geq0$ the chain of inequalities $$8^x+4^x=32^{x/2}\bigl(2^{x/2}+2^{-x/2}\bigr)\geq 30^{x/2}\bigl((6/5)^{x/2}+(6/5)^{-x/2}\bigr)=6^x+5^x\ .$$