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Albert1
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$m,n,k\in N$, and $m>1,n>1$
prove :
$(3^{m+1}-1)\times (5^{n+1}-1)\times(7^{k+1}-1)>98\times 3^m\times 5^n\times7^k$
prove :
$(3^{m+1}-1)\times (5^{n+1}-1)\times(7^{k+1}-1)>98\times 3^m\times 5^n\times7^k$
nice solution!kaliprasad said:becauase m > 1and n > 1
$(3^{m+1}-1)\times (5^{n+1}-1)\times(7^{k+1}-1)$
= $3^m(3- \frac{1}{3^m})\times 5^n(5-\frac{1}{5^n})\times 7^k(7- \frac{1}{7^k})$
= $3^m\times 5^n \times 7^k (3- \frac{1}{3^m})(5-\frac{1}{5^n})(7- \frac{1}{7^k})$
$\ge \ 3^m\times 5^n \times 7^k (3- \frac{1}{3^2})(5-\frac{1}{5^2})(7- \frac{1}{7})$ putting minimum values of m,n,k
$\ge 98.25 \times 3^m\times 5^n \times 7^k$ (used a calculator)
$\gt 98 \times 3^m\times 5^n \times 7^k$
An inequality is a mathematical statement that compares two quantities, indicating that one is greater than, less than, or equal to the other.
When m, n, and k are in the set of natural numbers, it means that they are positive integers (whole numbers) including zero.
When m and n are greater than 1, it means that they are both larger than the number 1. This also implies that they are positive integers.
To prove an inequality, you must use mathematical properties and operations to manipulate the given quantities and show that one side is greater than the other.
Yes, for example, we can prove that for any natural numbers m and n, if m > 1 and n > 1, then m + n > m. We can start by adding 1 to both sides of the inequality, giving us m + 1 > m + n. Then, we can subtract m from both sides, leaving us with 1 > n. Since n is a natural number and we know that 1 is the smallest natural number, this statement is true. Therefore, we have proven the inequality.