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Let $s,t\geq 0$ s.t. $s+t\geq 1$. Let $x_k,y_k\geq 0$. I'm trying to prove that $$\sum_{k=m}^n x_k^sy_k^t\leq \left(\sum_{k=m}^n x_k\right)^s\left(\sum_{k=m}^n y_k\right)^t.$$

Attempts

I remarked that if $s+t=1$ then $\frac{1}{1/s}+\frac{1}{1/t}=1$ and thus it's just Holder's inequality. But if $s+t\geq 1$, I don't know how to use Holder inequality. Therefore, it should has something else...

Michael Rozenberg
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user352653
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  • If $s$ and $t$ are integers, you can just note that every term of the form $x_k^s y_k^t$ appears in the expansion of the RHS, along with a bunch of other terms that must be positive. I don't know how this could be generalized to rationals, though. – Connor Harris Oct 27 '17 at 15:58

2 Answers2

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Consider $s',t'$ defined by $s' \stackrel{\rm def}{=}\frac{s}{s+t}$, $t' \stackrel{\rm def}{=}\frac{t}{s+t}$. Applying Hölder's inequality, $$ \sum_{k} x_k^{s'}y_k^{t'} \leq \left( \sum_{k} x_k \right)^{s'} \left( \sum_{k} y_k \right)^{t'} $$ so that $$ \left( \sum_{k} (x_k^{s}y_k^{t})^{\frac{1}{s+t}} \right)^{s+t} \leq \left( \sum_{k} x_k \right)^{s} \left( \sum_{k} y_k \right)^{t}\,.\tag{1} $$ Now, since $s+t\geq 1$, we have for non-negative $\alpha_k$'s that $$ \left( \sum_k \alpha_k\right)^{\frac{1}{s+t}} \leq \sum_k \alpha_k^{\frac{1}{s+t}} $$ (this is by monotonicity of $\ell_p$ norms, even though here we have a semi-norm) and thus $$ \sum_{k} x_k^{s}y_k^{t} \leq \left( \sum_{k} (x_k^{s}y_k^{t})^{\frac{1}{s+t}} \right)^{s+t}\tag{2} $$ and combining $(1)$ and $(2)$ gives the result.

Clement C.
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  • Let me know if I need to elaborate on the "monotonicity of $\ell_p$ norms" part. – Clement C. Oct 27 '17 at 15:55
  • I also strongly suggest that you check out (for more complicated or futures inequalities of that ilk to prove) to check the automated inequality prover of Gregory and Paul Valiant (paper, and accompanying Matlab code). Not only will it tell you whether the inequality holds, it will also output a proof (how to show it in few steps). http://theory.stanford.edu/~valiant/pap_link.html – Clement C. Oct 27 '17 at 16:11
  • Indeed, could you please develop monotonicity $\ell_p$ norm ? It's not Jensen, right ? (it would be the inverse inequality, $(\sum_{k} x_k)^{s+t}\leq\sum_{k}x_k^{s+t}$.) Thank you :) – user352653 Oct 29 '17 at 19:16
  • @user352653 See e.g. this. – Clement C. Oct 29 '17 at 19:51
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Let $s=pl$ and $t=ql$, where $l>0$ and $p+q=1$.

Thus, $l\geq1$ and by Holder $$\left(\sum_{k=m}^n x_k\right)^s\left(\sum_{k=m}^n y_k\right)^t=\left(\left(\sum_{k=m}^n x_k\right)^p\left(\sum_{k=m}^n y_k\right)^q\right)^l\geq\left(\sum_{k=m}^nx_k^py_k^q\right)^l.$$ Thus, it remains to prove that $$\left(\sum_{k=m}^nx_k^py_k^q\right)^l\geq\sum_{k=m}^n x_k^{pl}y_k^{ql}.$$ Now, let $x_l^py_k^q=a_k$.

Thus, we need to prove that $$\left(\sum_{k=m}^na_k\right)^l\geq\sum_{k=m}^na_k^l,$$ which is true by Karamata for the convex function $f(x)=x^l.$

Indeed, let $a_m\geq a_{m+1}\geq...\geq a_n.$

Thus, $$(a_m+a_{m+1}+...+a_n,0,...,0)\succ(a_m,a_{m+1},...,a_n)$$ and we are done!

Michael Rozenberg
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