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

Question

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)

Answer 1

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 $$3^{-t}=\bigl(3^{1/4}\bigr)^{-4t}\geq\left(1+{1\over3}\right)^{-4t}\geq1-{4\over3}t\qquad(t\geq0)\ .$$

Answer 2

I know you don't want analysis but this application of the mean value Theorem shouldn't be not posted:

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.