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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.

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  • $\begingroup$ Have you tried induction?(I don't think it will be very nice, but it could be effective). $\endgroup$ – S.C.B. Mar 4 '16 at 7:49
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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 $$3\frac{1^3+2^3+3^3}{1^2+2^2+3^2} - 2 \frac{1^3+2^3}{1^2+2^2} =3\frac{36}{14}-2\frac{9}{5}\approx 4.114$$

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