Prove the inequality $\frac{k^2}{e^k}\le \frac{4}{k+1}$

Question

I've been trying to prove that this inequality: $\frac{k^2}{e^k}\le \frac{4}{k+1}$ holds for all natural numbers. I believe the proof can be done using mathematical induction, but can't find a way how to do it.

Can anyone help me?

Answer 1

$$f(x)=k^2(k+1)e^{-k}<2k^3e^{-k}$$ Now find the maximum of $2k^3e^{-k}$ by taking the derivative and comparing to zero: $$3k^2e^{-k}=k^3e^{-k}\ \ \longrightarrow\ \ k=3$$ Thus, after verifying this extrema is indeed a maximum, we have that: $$f(x)<2\cdot (3/e)^3 <2\cdot (30/27)^3=\frac{2000}{729}<3$$ In actuality, $f(x)<1.8$, but that may be harder to prove.

Answer 2

Here's a proof without induction (but assuming you know the Taylor series for the exponential function).

The desired inequality is equivalent to proving $e^k\ge{1\over4}(k+k^3)$ for $k\in\mathbb{N}$. Now

$$e^k=1+k+{1\over2}k^2+{1\over6}k^3+{1\over24}k^4+\cdots$$

Since all terms are non-negative, we can truncate at $k^4$, so it suffices to prove

$$1+\left(1-{1\over4}\right)k+{1\over2}k^2+\left({1\over6}-{1\over4}\right)k^3+{1\over24}k^4\ge0$$

for $k\in\mathbb{N}$. On multiply through by $24$, this can be rewritten as

$$k^4-2k^3+12k^2+18k+24\ge0$$

This inequality actually holds for all $k$, not just non-negative integers, since

$$k^4-2k^3+12k^2+18k+24=(k^4-2k^3+k^2)+2k^2+(9k^2+18k+9)+15\\ =k^2(k-1)^2+2k^2+9(k+1)^2+15$$

Answer 3

$$e^k \ge \frac{k^2(k+1)}{4}$$

• Base $k=1,2$: $$e^1 \ge 1 \ge \frac{1(1+1)}{4}$$ $$e^2 \ge 7 \ge \frac{4(2+1)}{4}$$

• Induction step:

$$e^{k+1} \ge ee^{k}\ge e\cdot\frac{k^2(k+1)}{4} \ge \frac{(k+1)^2(k+2)}{4}$$

To ensure the last inequality, see that

$$ek^2\ge (k+1)(k+2)$$ for $k>2$.

Answer 4

We need to prove that $4e^k\geq k^3+k^2$, for which it's enough to prove that $f(x)\geq0$ for reals $x\geq1$, where $$f(x)=4e^x-x^3-x^2.$$ Indeed, $f'''(x)=4e^x-6\geq4e-6>0$.

Hence, $f''(x)=4e^x-6x-2\geq4e-8>0$.

Hence, $f'(x)=4e^x-3x^2-2x\geq4e-5>0$ and $f(x)\geq f(1)=4e-2>0$ and we are done!