# Prove that $3^x+1\geq 4x$ for $x \in (0,1]$

Prove that $$3^x+1\geq 4x$$ for $$x \in (0,1]$$.

I tried using Bernoulli’s inequality for real exponents but then I needed to prove an inequality that turned out to be false.

Please don’t use any limits or analysis, and prove all the inequalities you use (unless they are well-known as AM-GM, Bernoulli, Jensen and so on)

• The easiest way to prove such things is to use analysis. Note that $3^x$ is always bigger than $1$ for $0<x<1.$ So the inequality is automatically true for $x\le 1/2.$ Mar 21, 2020 at 12:17
• Not sure why $x>0$ is required, the inequality holds for $x \leq 1$
– Sil
Mar 21, 2020 at 13:02
• Yeah. But I need this inequality in a problem that only uses x in (0,1), that’s why I didn’t say x <=1 too. Mar 21, 2020 at 16:13

I'm using the Bernoulli inequality in the form $$(1+u)^v\geq 1+uv\qquad(u>-1, \quad v\leq0\ \ \vee \ v\geq1)\ .\tag{1}$$ Put $$x=1-t\quad(0\leq t<1)\ .$$ Then we have to prove that $$3^{1-t}+1\geq4-4t$$, or $$3^{-t}\geq1-{4\over3}t\qquad(0\leq t<1)\ .$$ From $$3<{256\over81}=\left({4\over3}\right)^4$$ it follows that $$3^{1/4}<{4\over3}$$. Using $$(1)$$ we therefore have $$3^{-t}=\bigl(3^{1/4}\bigr)^{-4t}\geq\left(1+{1\over3}\right)^{-4t}\geq1-{4\over3}t\qquad(t\geq0)\ .$$
Since $$\frac{\mathrm d}{\mathrm dx} 3^x = 3^x\ln(3)$$, we know by the mean value Theorem that for all $$x\in]0,1]$$ and some $$\xi=\xi(x)\in]x,1[$$ we have $$\frac{3^x-3}{x-1}=3^\xi \ln(3)\le 3\ln(3).$$
It follows by multiplying with $$x-1< 0$$ that
$$3^x-3\geq 3 \ln(3) (x-1)$$ so that $$3^x+1\geq 4+3 \ln(3) (x-1) = 4x+(x-1) (\ln (27)-4)>4x.$$
Since we have equality for $$x=1$$ as easily checked, the inequality follows.