TITLE: How to prove this inequality
QUESTION [5 upvotes]: To following problem,   I can't solve it unfortunately. Prove that for all integer values $n,p,q>1(p>q)$,, 
$$\dfrac{p}{q}(n+1)^{\frac{p}{q}-1}\ge (n+1)\cdot\left(\dfrac{1^p+2^p+\cdots+(n+1)^p}{1^q+2^q+\cdots+(n+1)^q}\right)-n\cdot\left(\dfrac{1^p+2^p+\cdots+n^p}{1^q+2^q+\cdots+n^q}\right)$$also I've tried to simplify that expression and I've found that it's equal to this , but I can't move on after that.

REPLY [2 votes]: You cannot prove it because it is wrong! Take $n=2, p=3, q=2$. Then obviously $n,p,q>1\;$ and $p>q.$ 
Now the LHS is
$\frac{3}{2}\sqrt{3}\approx 2.598$ but the RHS is 
$$3\frac{1^3+2^3+3^3}{1^2+2^2+3^2} - 2 \frac{1^3+2^3}{1^2+2^2}
=3\frac{36}{14}-2\frac{9}{5}\approx 4.114$$