$=1+\frac{\sum_{i=1}^{n}a^2_{i}-b^2_{i}}{\sum_{i=1}^{n}b_{i}(a_{i}+b_{i})}$. WLOG $b_1$ is minimal. $\sum_{i=1}^n a^2_{i}-b^2_{i} \le (\sum_{i=1}^n a_i)^2 - \sum_{i=1}^n b^2_i= (\sum_{i=1}^n b_i)^2 - \sum_{i=1}^n b^2_i$ $\sum_{i=1}^n b_i(a_i+b_i) \ge b_1(\sum_{i=1}^n a_i)+\sum_{i=1}^n b^2_i= b_1(\sum_{i=1}^n b_i)+\sum_{i=1}^n b^2_i$ If $\frac{\sum_{i=1}^{n}a^2_{i}-b^2_{i}}{\sum_{i=1}^{n}b_{i}(a_{i}+b_{i})} \le 0$, then $1+\frac{\sum_{i=1}^{n}a^2_{i}-b^2_{i}}{\sum_{i=1}^{n}b_{i}(a_{i}+b_{i})} \le 1 \le 1+\frac{(\sum_{i=1}^n b_i)^2 - \sum_{i=1}^n b^2_i}{ b_1(\sum_{i=1}^n b_i)+\sum_{i=1}^n b^2_i}$. If $\frac{\sum_{i=1}^{n}a^2_{i}-b^2_{i}}{\sum_{i=1}^{n}b_{i}(a_{i}+b_{i})} \ge 0$, then $1+\frac{\sum_{i=1}^{n}a^2_{i}-b^2_{i}}{\sum_{i=1}^{n}b_{i}(a_{i}+b_{i})} \le 1+\frac{(\sum_{i=1}^n b_i)^2 - \sum_{i=1}^n b^2_i}{ b_1(\sum_{i=1}^n b_i)+\sum_{i=1}^n b^2_i}$ Regardless, $1+\frac{\sum_{i=1}^{n}a^2_{i}-b^2_{i}}{\sum_{i=1}^{n}b_{i}(a_{i}+b_{i})} \le 1+\frac{(\sum_{i=1}^n b_i)^2 - \sum_{i=1}^n b^2_i}{ b_1(\sum_{i=1}^n b_i)+\sum_{i=1}^n b^2_i}$. Let $M=b_1$, $X=\sum_{i=2}^n b_i$, $Y=\sum_{i=2}^n b^2_i$, and $(n-1)M \le X$. Then $1+\frac{(\sum_{i=1}^n b_i)^2 - \sum_{i=1}^n b^2_i}{ b_1(\sum_{i=1}^n b_i)+\sum_{i=1}^n b^2_i}=1+\frac{(M+X)^2-(M^2+Y)}{M^2+MX+M^2+Y}$ $=1+\frac{2MX+X^2-Y}{2M^2+MX+Y}=\frac{2M^2+3MX+X^2}{2M^2+MX+Y}$. If we keep $M$ and $X$ fixed, this is a decreasing function of $Y$. Hence, it attains its maximum when $Y$ is minimum. By QM-AM, $Y \ge \frac{X^2}{n-1}$ Thus, this expression is $\le \frac{2M^2+3MX+X^2}{2M^2+MX+\frac{X^2}{n-1}} = (n-1) \frac{2M^2+3MX+X^2}{2(n-1)M^2+(n-1)MX+X^2}$. Because $n \ge 4$, the denominator of the second term is $\ge$ the numerator of the second term, so this is all $\le (n-1)$. Equality can be attained when $a_1=n-1$, $a_2=a_3=...=a_n=0$, $b_1=0$, $b_2=b_3=...=b_n=1$. Cheers, Rofler

things are different ...and difficult when n=3

ppwwyyxx wrote: things are different ...and difficult when n=3 Why is it different for $n=3?$

Why the question here?

SCP wrote: ppwwyyxx wrote: things are different ...and difficult when n=3 Why is it different for $n=3?$ for n=3, the answer is not n-1=2

ppwwyyxx wrote: SCP wrote: ppwwyyxx wrote: things are different ...and difficult when n=3 Why is it different for $n=3?$ for n=3, the answer is not n-1=2 It is then too difficult to calculate??

Not really. Now you only need to maximize $\frac{2M^{2}+3MX+X^{2}}{2M^{2}+MX+\frac{X^{2}}2}$ (see Rofler's solution). Since it is homogeneous, we can suppose $X = 1$. Notice that $M \leq 1/2$ (read Rofler's solution again), so we need to maximize the one-variable function \[f(M) = \frac{2M^{2}+3M+1}{2M^{2}+M+\frac12}\] in the interval $[0,1/2]$. This can be done with some Calculus (by taking the derivative), and you find that the maximum is attained at $M = \frac{\sqrt3 - 1}4$. The maximum value is $f\left(\frac{\sqrt3 - 1}4\right) = \frac{2\sqrt 3 + 3}3$. Notice that this value is bigger than $2$. Just one final observation: when $n \geq 4$, the roots of the derivative are not positive. In fact, the function is decreasing, so we set $M = 0$, which is the equality case for $n\geq 4$.

Excuse me for opening again this topic, but I've just seen this inequality and found different solution, so I want to post it. I dedicate this solution to the one girl named Lucija. She will know why. We will prove that for $n\ge 4$ we have: $\frac{\sum_{i=1}^n a_i(a_i+b_i)}{\sum_{i=1}^n b_i(a_i+b_i)} \leq n-1$ WLOG we can suppose $a_1 \ge ... \ge a_n$, and $\sum_{i=1}^{n} a_i=\sum_{i=1}^{n} b_i=1$. Let's denote $A=\sum_{i=1}^{n} a_i^2$ , $B=\sum_{i=1}^{n} b_i^2$ and $X=\sum_{i=1}^{n} a_ib_i$ We can see that we can always choose numbers such that $A> B$. Now we define: $f(X)=\frac{A+X}{B+X}$, so $f'(X)= \frac{B-A}{(B+X)^2}<0$ what means that we want to minimize $X$ to maximize fraction. Under the assumption, form rearrangement inequality we know that $X$ is minimal when $b_1 \le ... \le b_n$. Now let's fix $b_1,...,b_n$ and define: $F(a_1,...,a_n)= \frac{A+X}{B+X}$. We will prove that $F(a_1,a_2,...,a_n)\leq F(a_1+a_2,0,...,a_n)$ But this is obviously true since $a_1^2+a_2^2 \leq (a_1+a_2)^2$ and $a_1b_1+a_2b_2 \ge (a_1+a_2)b_1$. (Because we increase $A$ and decrease $X$). We can see that we can repeat this step $n-1$ times, so we conclude: $F(a_1,...,a_n) \leq F(1,0,...,0)$. It remains to prove: $F(1,0,...,0)=\frac{1+b_1}{\sum_{i=1}^{n} b_i^2 +b_1} \leq n-1$ From Cauchy-Schwarz we have: $\frac{1+b_1}{\sum_{i=1}^{n} b_i^2 +b_1} \leq \frac{1+b_1}{b_1^2 +\frac{(1-b_1)^2}{n-1}+b_1}$ And finally: $ \frac{1+b_1}{b_1^2 +\frac{(1-b_1)^2}{n-1}+b_1} \leq n-1 \; \Leftrightarrow b_1(nb_1+n-4) \ge 0$. What clearly holds for $n \ge 4$. Equality holds when $a_1=1, \; a_2=...=a_n=0$, and $b_1=0, \; b_2=...=b_n= \frac{1}{n-1}$

sorry for back a old topic,i solved it some days ago,since i am boring now,i will post my solution it is not hard to guess for $n\ge 4$,we have \[\frac{{\sum\limits_{i = 1}^n {{a_i}({a_i} + {b_i})} }}{{\sum\limits_{i = 1}^n {{b_i}({a_i} + {b_i})} }} \le n - 1\] or \[\left( {n - 1} \right)\sum\limits_{i = 1}^n {b_i^2} + \left( {n - 2} \right)\sum\limits_{i = 1}^n {{a_i}{b_i}} \ge \sum\limits_{i = 1}^n {a_i^2} \] or \[\left( {n - 1} \right)\left[ {\left( {n - 1} \right)\sum\limits_{i = 1}^n {b_i^2} + \left( {n - 2} \right)\sum\limits_{i = 1}^n {{a_i}{b_i}} -\sum\limits_{i = 1}^n {a_i^2} } \right] \ge 0\] but it is equivalent to \[\sum\limits_{i = 1}^n {{{\left[ {\left( {n - 1} \right){b_i} + {a_i}} \right]}^2}} + \left( {n - 1} \right)\left( {n - 4} \right)\sum\limits_{i = 1}^n {{a_i}{b_i}} \ge n\sum\limits_{i = 1}^n {a_i^2} \] since \[\left( {n - 1} \right)\left( {n - 4} \right)\sum\limits_{i = 1}^n {{a_i}{b_i}} \ge 0\] and \[\sum\limits_{i = 1}^n {{{\left[ {\left( {n - 1} \right){b_i} + {a_i}} \right]}^2}} \ge \frac{1}{n}{\left( {\sum\limits_{i = 1}^n {\left[ {\left( {n - 1} \right){b_i} + {a_i}} \right]} } \right)^2} = n{\left( {\sum\limits_{i = 1}^n {{a_i}} } \right)^2} \ge n\sum\limits_{i = 1}^n {a_i^2} \] we are done. not hard to see the equality can hold.