# 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?

• That AM-GM simplification isn't going to help you, because for $x$ large enough (larger than about $8.5$, to be more precise), the $6^x$ term will dominate the $\sqrt{32}^x$ term. – Arthur Nov 28 at 14:01

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

• @Macavity I also got the same result as you. – The Demonix _ Hermit Nov 28 at 17:28
• @Macavity: I was trying to match three of the four bases of the power function of the inequality. So: $$4\cdot \frac 32\cdot \frac43=8\ ,$$ $$4\cdot \frac 32\cdot 1=6$$ $$4\cdot 1\cdot 1=4$$ and the remained product should be stronger than $5$, that one is $$4\cdot 1\cdot \frac 43=\frac {16}3>5 \ .$$ – dan_fulea Nov 28 at 17:30
• OK - that’s clear now. Thanks. – Macavity Nov 28 at 17:35
• @TheDemonix_Hermit Thanks for the comment, please always ask / insert it in such situations! (It is a good check also for me, some answers - as this one - are a ten seconds shot... So a double check by the community is the thing i want.) I hope the situation is explained and correct now. Since i am also following didactical activities, and need to teach so that errors are prevented by mnemonic mechanisms, please give a hint how you obtained the other four numbers. ($12$ was one of them, nine was - i think - also in the liist.) – dan_fulea Nov 28 at 17:56
• @dan_fulea Clearing denominators by multiplying by $6^x$ And expanding will give the other four numbers. The $4^x$ is cancelled as always positive. – Macavity Nov 28 at 18:29

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

• Dear downvoter, thank you so much for your useful comments and generous support. – Robert Z Nov 28 at 18:34

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

Note that $$8^x=(6^x)^{\log_6(8)}$$ so that $$$$\tag 1\label 14^x+8^x-6^x-5^x\geq 8^x-6^x-5^x\to\infty$$$$ as $$x\to\infty$$.

I will show that the only non-negative solution to $$$$\tag 2\label 28^x+4^x=6^x+5^x$$$$ is $$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 $$x>0$$.

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.

• How do you prove it for all real $x$, esp $x \in (0,1)$? – Calvin Lin Nov 29 at 9:18