Prove $\sum_{k=m}^nx_k^sy_k^t\leq \left(\sum_{k=m}^n x_k\right)^s\left(\sum_{k=m}^n y_k\right)^t$

Question

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...

Answer 1

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.

Answer 2

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!