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Inequality of cubic and exponential functions

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,597
Prove that $3^n\ge(n+3)^3$ for any natural number $n\ge6$.
 

kaliprasad

Well-known member
Mar 31, 2013
1,283

This problem is same as
$3^{n-3} > = n^3$ or $3^n >= 27n^3$ for $n >=9$
To prove the same we use principle of mathematical induction

Base step
For n = 9 $LHS = 3^9 = 3^3 * 3^6 = 27 * 9^3$ so $3^n >= 27n^3$
So base step is true
Now $(\frac{n+1}{n})^3$ decreases as n increases and at n = 9 we have $(\frac{n+1}{n})^3= \frac{1000}{729}< 3$
So $(\frac{n+1}{n})^3< 3$ for all $n>=9$
Or $3 > (\frac{k+1}{k})^3\cdots(1)$ for all $k>=9$

Induction step
Let it be true for n = k $k >=9$
We need to prove it to be true for n = k+ 1
$3^k > = 27 k^3$
Multiplying by (1) on both sides
$3^{k+1} > = 27 * (\frac{k+1}{k})^3 * k^3 $
Or $3^{k+1} >= 27(k+1)^3$
So it is true for n = k+ 1
We have proved the induction step also

Hence proved