Proving an inequality involving an exponential and a polynomial?

Question

$$106+2^{x}>2x^{2}+2x$$

I have tried to prove that this inequality is true for $x\ge 0$. I would be happy for any hint!

Thanks for the two answers! They are very helpful. But I am interested in a solution including induction.

Answer 1

It is easy to see that the both the RHS and LHS are monotonic increasing on the interval $[0,\infty)$ and that for $x\in[0,6]$ you have $2x^2+2x\leq 84<106<106+2^x$, implying that $2x^2+2x<106+2^x$ for all $x\in[0,6]$

From here, using calculus, consider the rate of change of each function.

The derivative of the LHS is $\frac{d}{dx}[106+2^x] = 2^x\ln 2$ and of the RHS is $\frac{d}{dx}[2x^2+2x]=4x+2$

At $x=6$ the derivative of the LHS is approximately $44$ and of the RHS is $26$. Showing that $2^x\ln 2 > 4x+2$ for all $x\geq 6$ is of similar difficulty to the original inequality, so let us continue to the second derivative.

Here we see that $\frac{d^2}{dx^2}[106+2^x] = 2^x(\ln 2)^2$ and $\frac{d^2}{dx^2}[2x^2+2x] = 4$. From here it is trivial to see that the second derivative is always larger on the interval $[6,\infty)$.

This implies that the derivative of the LHS is always larger on the interval $[6,\infty)$, implying then that the original function on the LHS is always larger on the interval $[6,\infty)$.

Combined with what we noticed before, we have that $106+2^x > 2x^2+2x$ for all $x\geq 0$

---

Using induction we can prove the statement true for natural numbers, but will require some more difficult arguments in order to extend the proof to be about the real numbers:

The proof is very similar: first we notice that for each $x\in\{0,1,\dots,6\}$ you have $2x^2+2x\leq 84 < 106 < 106+2^x$.

Suppose that the claim is true for some $x\geq 6$ that $2x^2+2x< 106+2^x$.

Then $\begin{array}{rll}2(x+1)^2+2(x+1) &= 2x^2+2x+4x+1&\text{using}~x+1~\text{instead}\\ &<106+2^x+4x+1&\text{by induction hypothesis}\\&<106+2^x+4x+4x&\text{since}~x>1\\&=106+2^x+8x&\text{by simplification}\\&<106+2^x+2^x&\text{since}~x>6~(\star)\\&=106+2^{x+1}&\text{by simplification}\end{array}$

Seeing that step $(\star)$ is true is again easiest with calculus, but suffice to say that it follows since $2^x$ grows much faster than $8x$. A short induction proof suffices: $8\cdot 6 = 48<64=2^6$. Supposing true for some $x\geq 6$ you have $8\cdot (x+1) = 8\cdot x + 8 < 8\cdot x+8\cdot x = 2\cdot 8\cdot x < 2\cdot 2^x = 2^{x+1}$

Thus we have shown that the desired inequality is true for all natural numbers $x\in\mathbb{N}$.

Answer 2

This is probably a too complex solution since based on fonction analysis.

Consider the function $$f(x)=-2 x^2-2 x+2^x+106$$ and its first and second derivatives $$f'(x)=-4 x-2+2^x \log (2)$$ $$f''(x)=2^x \log ^2(2)-4$$ The first derivative cancels at two points $$x_1=\frac{-2 W\left(-\frac{\log ^2(2)}{4 \sqrt{2}}\right)-\log (2)}{2 \log (2)}\approx -0.365494$$ $$x_2=\frac{-2 W_{-1}\left(-\frac{\log ^2(2)}{4 \sqrt{2}}\right)-\log (2)}{2 \log (2)}\approx 4.98399 \approx 5$$ $W(z)$ being Lambert function. We discard $x_1$ since negative.

The second derivative test shows that $x_2$ corresponds to a minimum and $f(5)=78$ (the exact value would be $f(x_2)\approx 77.9986$).

So, for any $x>0$, $f(x)>0$ and $2^x+106>2 x^2+2 x$.