Big And Small? Or Two Mediums?

We wrote $10^n+20^n = (15-5)^n+(15+5)^n$ . For any natural number $n$ , the value of $(x+y)^n+(x-y)^n$ will be $\large 2\sum_{i=0}^{\lceil n/2 \rceil}\binom{n}{2i}x^{n-2i}y^{2i} ≥ 2x^n$

Expand the binomial sigma would gives $2\binom{n}{0}x^n+2\binom{n}{2}x^{n-2}y^2+2\binom{n}{4}x^{n-4}y^4+\dots+2\binom{n}{2\lceil n/2 \rceil}x^{n-2\lceil n/2 \rceil}y^{2\lceil n/2 \rceil}$

Cancelling $2x^n$ on both sides give you the inequality at minimum of zero. Hence, $10^n+20^n ≥ 2(15^n)$

Trying $n=2$ we see that $10^2 + 20^2 > 15^2 + 15^2$ .

From the phrasing of the question, we can build the hypothesis that

$10^n + 20^n > 15^n + 15^n, n \in \mathbb{Z} \mid n > 1$

and try to prove this by induction.

Assuming that this statement is true for $n = k$ , for $n = k+1$ we have:

$10^{k+1} + 20^{k+1} = 10(10^k) + 20(20^k) = 10(10^k + 20^k) + 10(20^k)$ $15^{k+1} + 15^{k+1} = 15(15^k) + 15(15^k) = 15(15^k + 15^k) = 10(15^k + 15^k) + 5(15^k + 15^k)$

We now need to establish which of these is greater. From our assumption, $10(10^k + 20^k) > 10(15^k + 15^k)$ . Looking at the remaining parts of these results, namely comparing $10(20^k)$ and $5(15^k + 15^k)$ , since the latter simplifies to $10(15^k)$ and clearly $10(20^k) > 10(15^k)$ , we have that $10^{k+1} + 20^{k+1} > 15^{k+1} + 15^{k+1}$ . So, since if the hypothesis is true for $n = k$ it is also true for $n = k + 1$ , it is thus proven true for all integers $n$ greater than 1.

My solution is based on that:

Take $0\leq \alpha \leq 1$ and two other real numbers, here 10 and 20 are the numbers we are interested in.

For all real numbers $x$ , $10(1-\alpha)x+20{\alpha}x = 10(\alpha+1)x$

I will note $(\alpha,10,20) \iff (10(\alpha+1))$ (multiplying $100(1-\alpha)$ % of a real number $x$ by 10 and $100{\alpha}$ % of the same real number $x$ by 20 is equivalent to multiply $x$ by $10(\alpha+1)$ ).

We can see that for all integers $n\geq 2 , 15^{n+1}+15^{n+1}=15(15^n+15^n)$ and $10^{n+1}+20^{n+1}=10 \times 10^n + 20 \times 20^n$

So, if you consider the sequence $(u_n=10^n+20^n)_{n \geq 2}$ ,

$\forall n \geq 2,$ if you consider $x=u_n, (\frac{20^n}{10^n+20^n},10,20) \iff (10(\frac{20^n}{10^n+20^n}+1)=\gamma_n)$

And because $\forall n \geq 2, \frac{20^n}{10^n+20^n} = \frac{1}{\frac{1}{2^n}+1} \geq \frac{1}{\frac{1}{4}+1} = \frac{4}{5}$ , we have: $\gamma_n \geq 18>15$

And finally, because $10^2+20^2=500 > 450= 15^2+15^2$ , we can conclude that:

$\boxed{\forall n \geq 2, 10^n+20^n>15^n+15^n}$

Divide both sides by 5^n.

You get 2^n + 4^n and 3^n + 3^n.

Assuming they would be equal for all values of n, then the differences between 2^n and 3^n and 4^n and 3^n will sum to zero.

Only when n = 1 are the two expressions equal. As n increases, 4^n tends towards infinity quicker than 3^n which also tends to infinity quicker than 2^n, as each time you increase n by 1, you multiply by 4,3 & 2 respectively

Therefore (4^n - 3^n) + (2^n - 3^n) > 0 so 2^n + 4^n > 3^n + 3^n, which thereby proves the original.

I guess there was no need to divide by 5^n. I just thought it would help! Sorry for my unsophisticated language. This is just how I "saw" it!

Dividing both the numbers by 5^n, we get 4^n+2^n and 2*3^n. For all n=3 or more, we can write n=a+3. So, we have 4^(a +3)=64×4^a>54×3^a=2×3^3×3^a =2×3^(a+3). Therefore, 4^n>2×3^n. So, obviously, 4^n+2^n>2×3^n. For n=2 also it is true. Hence proved.