EDITI am almost sure ,there is not a single relationship between this problem and HOLDER inequality(beside of their looks ) ,I have checked my solution , and its right without any use of holder

There exist such numbers,$\alpha=\beta=1$ Any one cliaming to have a solution by holder ,Please waste your time for posting it.Thanks in advance

Omid Hatami wrote: Find all $\alpha>0$ and $\beta>0$ that for each $(x_1,\dots,x_n)$ and $(y_1,\dots,y_n)\in\mathbb {R^+}^n$ that:\[(\sum x_i^\alpha)(\sum y_i^\beta)\geq\sum x_iy_i\] I think it's hard and stronger than Holder inequatily and I think we most prove that $f(x)=(\sum x_i^\alpha)^{\frac {1}{\alpha}}$ is the increasing function and when $\frac {1}{\alpha}+\frac {1}{\beta} \geq 1$ so that Holder inequality should be right . Omid are you agree with me?

Wehave only to prove this but it is difficult $(\sum x_i^\alpha)(\sum y_i^\beta)\leq (\sum x_i^\alpha)^{\frac {1}{\alpha}} (\sum y_i^\beta)^{\frac {1}{\beta}}$

lomos' solution doesn't make sense to me. Anyways, say that there is an $\alpha$ and a $\beta$ such that the inequality holds for $(x_1,x_2,\ldots,x_n)$ and $(y_1,y_2,\ldots,y_n)$ Replace each $x_i,y_i$ with $tx_i,ty_i$. Then if the inequality still holds, we must have $t^{\alpha+\beta}\geq t^2$. If $t\geq 1$, then $\alpha+\beta\geq 2$. That means that one of $\alpha,\beta$ is $\geq 1$ (wlog, $\alpha$). Put $x_i=\epsilon$ for each $i$. The inequality becomes $\left(n\epsilon^{\alpha-1}\right)\left(\sum y_i^\beta\right)\geq \sum y_i$ If $\alpha>1$, then for any choice of the $y_i$'s and $\beta$, we can make the left hand side arbitrarily small, whereas the RHS is always positive. Otherwise, we need only consider the case where $\alpha=\beta=1$. But this case is obvious, since the expansion of the LHS contains the RHS and other strictly positive terms.

lomos_lupin wrote: $\alpha+\beta=2$ Yes, otherwise the inequality is dimensionally incorrect. Quote: So $\alpha=\beta=1$ is the only posiblity.which obviously is wrong When $\alpha = \beta = 1$ the inequality is $\sum x_i \sum y_i \geq \sum x_i y_i$, which is trivially true when the $x,y$ are positive.

Fleeting guess: Oops,my bad ,of course ,why did i write (the ineqaulity is obviously Wrong???) Quote: lomos solution doesnt make sense to me Yes ,some solution was not precise ,so i shall exaplain to make in right: Let $\alpha ,\beta$ be two numbers satisfying the condition of the problem so for any $2n$ positive numbers like $(x_1,...,x_n),(y_1,...,y_n)$,the inequality occurs . Put $(x_1,...,x_n)=(y_1,...,y_n)=(w,w,...,w)$ Where $w$ is an arbitarily positive number. then the ineqaulity becoms $w^{\alpha+\beta-2}\ge \frac1n$ But $w$ is an arbitarily real number and $n$ is a natural number ,so in the case that $\alpha+\beta \neq 2$ We can find a counterexample. so $\alpha+\beta=2$. Now put $(x_1,...,x_n)=(w,w,...,w)$ and $(y_1,...,y_n)=(w^2,...,w^2)$ so the ineqaulity becomes $w^{2\beta+\alpha-3} \ge \frac1n$ whith the same arguments like above we have $\alpha+2\beta=3$ So $\begin{cases} \alpha+\beta=2 \.1in] \alpha+2\beta=3\end{cases} \longrightarrow \alpha=\beta=1$ In this case $\alpha=\beta=1$ the inequality occurs ,so this is the only answer.

dear lomos_lupin,you are right,this problem is from Iran National Math Olympiad (3rd Round) 2005.but Omid Hatami had a typo mistake here.the original problem is: Quote: Find all $\alpha>0$ and $\beta>0$ that for each $(x_1,\dots,x_n)$ and $(y_1,\dots,y_n)\in\mathbb {R^+}^n$: \[(\sum x_i^\alpha)^{\frac{1}{\alpha}}(\sum y_i^\beta)^{\frac{1}{\beta}}{\geq\sum x_iy_i}\]