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Without using a calculator, show that 
, and for a positive integer 
, prove that
.
, and for a positive integer 
, prove that
.
Inequality with integers and indices
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#2
Mar 17, 2025, 1:09 PM
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In order to solve the first problem, I have the following idea: raise both sides to the 6th power, and you're done! Right?
In the general case, consider the function f(x)=lnx/x, whose derivative is negative for x>e, so the function is decreasing. Let x=n and x=n+1, and compare them! I think you understand, if not, write and I'll answer!
In the general case, consider the function f(x)=lnx/x, whose derivative is negative for x>e, so the function is decreasing. Let x=n and x=n+1, and compare them! I think you understand, if not, write and I'll answer!
This post has been edited 3 times. Last edited by invisibleman, Mar 18, 2025, 11:36 AM
Reason: typo
Reason: typo
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The inequality is equivalent to 
. Taking natural logarithm on both sides, this is equivalent to 
Dividing, this is equivalent to 
. Since 
is increasing (obviously linear functions increase more quickly than logarithms, or using the derivative like @above said), we are done by definition.
i hope this is right!
well someone taught me the solution hehe
. Taking natural logarithm on both sides, this is equivalent to 
Dividing, this is equivalent to 
. Since 
is increasing (obviously linear functions increase more quickly than logarithms, or using the derivative like @above said), we are done by definition.i hope this is right!
well someone taught me the solution hehe
This post has been edited 2 times. Last edited by krish6_9, Mar 18, 2025, 12:31 AM
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#6
Mar 18, 2025, 1:43 AM
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Bro! Let's show n^(n+1) > (n+1)^n , or (1+1/n)^n < n. Note that (1+1/n)^n < e because this limits upwards towards e. So for n >= 3, this is obviously true as e<3. Bruh! Now that's rizz!
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#7
Mar 18, 2025, 7:32 AM
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f(x)=\frac{\ln x}{x}\Rightarrow f'(x)=\frac{1-\ln x}{{{x}^{2}}}<0,\forall x>e\Rightarrow f(n)>f(n+1),\forall n\ge 2
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