Prove an inequality with sizable exponents

Question

The Problem

Show $2^{100} + 3^{100} \lt 4^{100}$

My Attempt

$$2^{100} + 3^{100} \lt 4^{100}$$

Rewriting $4^{100}$ as $2^{200}$

$$\implies 3^{100} \lt 2^{200} - 2^{100}$$

$$\implies 3^{100} \lt 2^{100}(2^{100} - 1)$$

As $2^{99} \lt 2^{100} - 1$, to prove our original statement it is sufficient to show $3^{100} \lt 2^{100}(2^{99})$

$$3^{100} \lt 2^{100}(2^{99})$$

$$\implies 3^{100} \lt 2^{199}$$

Making the power of 2 a nicer number

$$\implies 2(3^{100}) \lt 2^{200}$$

$$\implies 2(3^{100}) \lt 4^{100}$$

$$\implies 2 \lt \frac{4^{100}}{3^{100}}$$

$$\implies 2 \lt (\frac{4}{3})^{100}$$

I can show this by looking at smaller powers of the ratio $\frac{4}{3}$

$$(\frac{4}{3})^{3} = \frac{64}{27} \gt 2$$

$$\implies 2 \lt (\frac{4}{3})^{3} \lt (\frac{4}{3})^{100}$$

Thus the original inequality holds. QED

Questions

Is this prove rigorous enough?

Have I made any assumptions that are non-trivial and should be shown?

Is there a more elegant method?

Answer 1

This may seem nitpicky but be it is VERY important that when you start the proof with

$2^{100} + 3^{100} < 4^{100}$ that you state that you do NOT know that to be true but that that is your desired result.

And it is important that you don't say $2^{100} + 3^{100} < 4^{100} \implies 3^{100} < 2^{200} - 2^{100}$. That is claiming you are starting for a hypothesis (that you do not know) and are trying to conclude a result (that you do not want; you want the hypothesis).

What you should say is something like: " $2^{100} + 3^{100} < 4^{100}$ would be true if we could show $3^{100} < 2^{200} - 2^{100}$.

In other words, replace all $\implies$ with $\Leftarrow$ and .... presto, your proof works!

And it does. There is no error (other than the implication misdirection).

However it'd be better if you used this as a rough draft and worked forward not backwards.

$2 < (\frac 43)^3 < (\frac 43)^{100}\implies.... \implies 2^{100} + 3^{100} < 4^{100}$.

Although... now I'm getting into style critique... working forward yields other emphases and ... reveals a faster solution:

$(\frac 43)^3= \frac {64}{27} > 2$.

So $2 < (\frac 43)^3 < (\frac 43)^{100}$.

So $2*3^{100} < 4^{100}$

So $3^{100} + 3^{100} < 4^{100}$. As $1 < 2 <3$ we know $2^{100} < 3^{100}$.

So $2^{100} + 3^{100} < 3^{100} + 3^{100} <4^{100}$.

Answer 2

Because $$\left(\frac{1}{2}\right)^{100}+\left(\frac{3}{4}\right)^{100}<\frac{1}{2}+\left(\frac{27}{64}\right)^{\frac{100}{3}}<\frac{1}{2}+\frac{1}{2}=1.$$

Answer 3

Note that

$$2^{100} + 3^{100} < 3^{100}+ 3^{100}=2\cdot 3^{100}\lt 4^{100}\iff 2<\left(\frac43\right)^{3}<\left(\frac43\right)^{100}$$

Answer 4

As written, your proof is incorrect. The problem is that you started off with the inequality you want to prove and manipulated it until you found a true statement. This is logically incorrect and invalidates the proof. It's the same as writing

$$0 = 1 \implies 6 = 7$$ by adding $6$ to both sides; but since the hypothesis is wrong, there is nothing we can conclude.

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However, your steps are all reversible. So you can replace the forward implications with the reverse ones, and the idea works. But you've only done the scratchwork, not the proof.

Answer 5

For fun:

Show:

$ 2^{100} \lt 4^{100} -3^{100}.$

Consider

$f(x)= x^{100}$, $x \gt 0,$ real.

MVT:

$\dfrac{f(4)-f(3)}{1} =f'(a)$,

where $3\lt a \lt 4$, i.e.

$4^{100} -3^{100} = 100(a)^{99} =(100/a)a^{100}.$

Hence :

$4^{100} -3^{100} = (100/a)a^{100} \gt (100/4)3^{100} =$

$(25)3^{100} \gt 2^{100}.$

Used: $f(x) =x^{100}$ is an increasing function.