This one was proposed by me

Inequality 16 in http://rho.math.uni-rostock.de/SemSkripte/Skript_Peter1.pdf with $r=-1$.

pi_quadrat_sechstel wrote: Inequality 16 in http://rho.math.uni-rostock.de/SemSkripte/Skript_Peter1.pdf with $r=-1$. emm,but not English

Nice problem!!

How long did it take you to type that LaTeX though...

DNCT1 wrote: Nice problem!!

Vietnam NMO 1962 : Prove that for positive real numbers $ a$, $ b$, $ c$, $ d$, we have\[ \frac{1}{\frac{1}{a}+\frac{1}{b}}+\frac{1}{\frac{1}{c}+\frac{1}{d}}\le\frac{1}{\frac{1}{a+c}+\frac{1}{b+d}}\] The inequality is equivalent to \[ \frac{ab}{a+b}+\frac{cd}{c+d}\le \frac{(a+c)(b+d)}{a+b+c+d}\] \[\left (\frac{a+b}{4}-\frac{ab}{a+b} \right )+\left (\frac{c+d}{4}-\frac{cd}{c+d} \right )\ge \frac{a+b+c+d}{4}-\frac{(a+c)(b+d)}{a+b+c+d}\] \[\frac{(a-b)^2}{a+b}+\frac{(c-d)^2}{c+d}\ge\frac{(a+c-b-d)^2}{a+b+c+d}\] Which is obviously true by Cauchy-Schwarz.

The offical solution uses induction on $n$.

Minkowski Inequality？

\begin{align*} &\sum_{i = 1}^m \left( \sum_{j = 1}^n \frac{1}{a_{ij}} \right)^{-1} \stackrel{Titu}{\le} \sum_{i=1}^m \left( \frac{n^2}{\sum_{j=1}^n a_{ij}}\right)^{-1}= \frac{1}{n^2} \sum_{i=1}^m \sum_{j=1}^n a_{ij} \\ &\left( \sum_{j = 1}^n \left( \sum_{i = 1}^m a_{ij} \right)^{-1} \right)^{-1} = \left(\sum_{j=1}^n \frac{1}{\sum_{i=1}^m a_{ij}}\right)^{-1} \stackrel{AM-GM}{\ge} m\left(\sum_{j=1}^n \frac{1}{\sqrt[m]{\prod_{i=1}^m a_{ij}}}\right)^{-1} \\ &WLOG \ \sum_{i=1}^m \sum_{j=1}^n a_{ij}= n^2 \end{align*} Now we have not so hard differentiable expressions, we can just lagrange bash. From above inequalities, we have to prove: $$ m \ge \sum_{j=1}^n \frac{1}{\sqrt[m]{\prod_{i=1}^m a_{ij}}}$$ Where $ \sum_{i=1}^m \sum_{j=1}^n a_{ij}= n^2$ Differentiating the contraint in terms of any variable $a_{ij}$ gives $1$. Hence we have to solve system of equations: $$\frac{d \left( \sum_{j=1}^n \frac{1}{\sqrt[m]{\prod_{i=1}^m a_{ij}}}\right)}{d(a_{1,1})}=...=\frac{d \left( \sum_{j=1}^n \frac{1}{\sqrt[m]{\prod_{i=1}^m a_{ij}}}\right)}{d(a_{ij})} $$ The expressions can look complex, but, differentiating for all $a_{i_0,1}$: $$ \frac{d \left( \sum_{j=1}^n \frac{1}{\sqrt[m]{\prod_{i=1}^m a_{ij}}}\right)}{d(a_{i_0,1})}= \frac{d \left( \frac{1}{\sqrt[m]{a_{i_0,1}}} \cdot \frac{1}{\sqrt[m] {\prod_{i=1,i \neq i_0}^m a_{i,1}}}\right)}{d(a_{i_0,1})}= -\frac{a_{i_0,1}^{\frac{-(m+1)}{m}}}{m} \cdot \frac{1}{\sqrt[m] {\prod_{i=1, i \neq i_0}^m a_{i,1}}} $$ Now, differentiating for all $a_{i,1}$ will not change value of $m$. therefore we can cancel the power of $m$ stuff, the $m$ in dominator, the $m$ root. Then the stuff gets pretty simple. for example, differentiate for $a_{i_1,1}$ again and find when they are equal: $$ \frac{a_{i_0,1}}{\prod_{i=1,i \neq i_0}^m a_{i,1}}=\frac{a_{i_1,1}}{\prod_{i=1,i \neq i_1}^m a_{i,1}}\ \Rightarrow a_{i_0.1}^2=a_{i_1,1}^2$$ Since variables can't be negative, we have $a_{i_0,1}=a_{i_1,1}$. Repeating for other $a_{i,1}$, we get $a_{1,1}=a_{2,1}=a_{3,1}=...=a_{m,1}$. We can do the same things for $j=2,3,...,n$. Now, differentiate similiarly for all $a_{1,j_0}$. We will get the similiar result but $a_{1,1}=a_{1,2}=...=a_{1,3}=a_{1,n}$ and so on. Since two of the results share one same variable, we will get all variables $a_{ij}$ must be equal for the minimum value. ( Obviously it is not the maximum value.). Now it is easy to finish since by contraint all $a_{ij}= \frac{n}{m}$ and equalty indeed holds. Therefore, we have proved that equality is right and equality case is when $a_{1,1}=...=a_{ij} $ $\blacksquare$

pi_quadrat_sechstel wrote: Inequality 16 in http://rho.math.uni-rostock.de/SemSkripte/Skript_Peter1.pdf with $r=-1$. There might be some problem with your link. Could you please check it and post again? thx