Prove Inequality: $m,n,k\in N$, $m>1,n>1$

[Albert1]

$m,n,k\in N$, and $m>1,n>1$
prove :
$(3^{m+1}-1)\times (5^{n+1}-1)\times(7^{k+1}-1)>98\times 3^m\times 5^n\times7^k$

 

[kaliprasad]

Spoiler

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)
$\gt 98 \times 3^m\times 5^n \times 7^k$

 

[Albert1]

Spoiler

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)
$\gt 98 \times 3^m\times 5^n \times 7^k$

nice solution!