The inequality $(3^{m+1}-1)\times (5^{n+1}-1)\times(7^{k+1}-1)>98\times 3^m\times 5^n\times7^k$ is proven for natural numbers $m, n, k$ where $m > 1$ and $n > 1$. The proof utilizes properties of exponential growth and the specific values of the bases involved. Key steps include demonstrating the growth rates of the left-hand side compared to the right-hand side, confirming that the inequality holds true for all specified values of $m$, $n$, and $k$.
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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)
$\gt 98 \times 3^m\times 5^n \times 7^k$