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

Question

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?

Answer 1

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

Answer 2

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 $x>0$.

Answer 3

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

Answer 4

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

Answer 5

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 ,

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

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.