It suffices to prove that \[ \sum_{i = 1}^n \frac{1}{a_i + 1} > 6 \]This should be immediate by CS, but since $a_i$s are positive integers, by smoothing argument, they should be minimized when all of them are $2$ or $3$. Doing simple equation stuff gives us \[ LHS \ge \frac{1}{3} \cdot 7 + \frac{1}{4} \cdot 15 > 6 \]

I smell some convexity, thought not sure.

We need to prove $$ \sum_{i = 1}^{22} \frac{1}{a_i + 1} > 6 $$Let $f(a_1,...,a_{22})= \sum_{i = 1}^{22} \frac{1}{a_i + 1}$ Let $a>b+1$ are two positive integers. Then $\frac{1}{a+1}+\frac{1}{b+1}-\frac{1}{a}-\frac{1}{b+2}=(a+b+2) \frac{a(b+2)-(a+1)(b+1)}{a(a+1)(b+1)(b+2)}= \frac{(a+b+2)(a-b-1)}{a(a+1)(b+1)(b+2)} > 0$ So if among $a_1,a_2,...,a_{22}$ there are two values $a_i,a_j$ such that $a_i-a_j \geq 2$ then $f(a_1,...,a_{22})-f(a_1,...a_i-1,...,a_j+1,...,a_{22})>0$. So minimal value of $f(a_1,...,a_{22})$ will be if $|a_i-a_j| \leq 1$ or $a_1=a_2=...=a_i=a,a_{i+1}=a_{i+2}=...=a_{22}=a+1$ and $ai+(22-i)(a+1)=59$ $22a-37=i \to \frac{38}{22}\leq a \leq \frac{59}{22} \to a=2,i=7$ and so $f(a_1,...,a_{22}) \geq 7* \frac{1}{3}+15\frac{1}{4}= 6+\frac{1}{12}$

Tintarn wrote: Let $a_1,a_2,\dots,a_{22}$ be positive integers with sum $59$. Prove the inequality \[\frac{a_1}{a_1+1}+\frac{a_2}{a_2+1}+\dots+\frac{a_{22}}{a_{22}+1}<16.\]

We have the function \[f(x) = \frac{x}{x+1}\]. Whose second derivative is easy to see that it is \[f''(x)=\frac{-2}{(x+1)^3}\]which is clearly negative for all $x > 0$ ,thus it is concave and we can apply Jensen's inequality. Where \[\sum_{n=1}^{22}\frac{a_n}{a_n+1}\leq 22f(\frac{59}{22})\]. Since the RHS is 16.02... but equality only happens when all equal but $22 \nmid 59$, it follows that the sum is less.

FibonacciMoose wrote: Since the RHS is 16.02... but equality only happens when all equal but $22 \nmid 59$, it follows that the sum is less. Less than $16.02$, yes. But why less than $16$? (You are close of course, but not quite there.)

sqing wrote: Tintarn wrote: Let $a_1,a_2,\dots,a_{22}$ be positive integers with sum $59$. Prove the inequality \[\frac{a_1}{a_1+1}+\frac{a_2}{a_2+1}+\dots+\frac{a_{22}}{a_{22}+1}<16.\] 我恰好看懂

For every $1\leqslant i<j\leqslant 22$, if $a_j-a_i\geqslant 2$, then $\dfrac{a_i}{a_i+1}+\dfrac{a_j}{a_j+1}\leqslant\dfrac{a_i+1}{a_i+2}+\dfrac{a_j-1}{a_j}$. So we can let $a_1=a_2=\cdots a_k=x,a_{k+1}=a_{k+2}=\cdots =a_{22}=x+1$, it is easy to see that $x=2,k=7$ $\therefore\dfrac{a_1}{a_1+1}+\dfrac{a_2}{a_2+1}+\cdots+\dfrac{a_{22}}{a_{22}+1}\leqslant\frac{2}{3}\times 7+\frac{3}{4}\times 15=\frac{191}{12}<16\Box$

EthanWYX2009 wrote: For every $1\leqslant i<j\leqslant 22$, if $a_j-a_i\geqslant 2$, then $\dfrac{a_i}{a_i+1}+\dfrac{a_j}{a_j+1}\leqslant\dfrac{a_i+1}{a_i+2}+\dfrac{a_j-1}{a_j}$. So we can let $a_1=a_2=\cdots a_k=x,a_{k+1}=a_{k+2}=\cdots =a_{22}=x+1$, it is easy to see that $x=2,k=7$ $\therefore\dfrac{a_1}{a_1+1}+\dfrac{a_2}{a_2+1}+\cdots+\dfrac{a_{22}}{a_{22}+1}\leqslant\frac{2}{3}\times 7+\frac{3}{4}\times 15=\frac{191}{12}<16\Box$ nice solution!

I think that this is problem is really cool for an inquality. Tlt in its original form is too weak, but as we only need the line to lie above the the curve $f(x)=\frac{x}{x+1}$ for integer values of $x$ and since it is intuitive to guess that the inequality is almost an equality if most of the terms are equal to $2$ or $3$, we consider the secant through the points on the curve at $x=2$ and $x=3$, which is given by $g(x)=\frac{x}{12}+1/2$, and for which it is easy to check that $g(x)\ge f(x)$ for $x\ge 3$ and $x\le 2$. Summing this gives a valid solution.

Let $a_1,a_2,\dots,a_{22}$ be positive integers with sum $59$. Prove the inequality \[\frac{a_1}{a_1+1}+\frac{a_2}{a_2+1}+\dots+\frac{a_{22}}{a_{22}+1}\le \frac{191}{12}\]

The inequality is equal to: $\sum_{i=1}^{22}\frac{1}{a_i+1}>6$ Let $f(a_1,\dots,a_{22})=\sum_{i=1}^{22}\frac{1}{a_i+1}, g(a_1,\dots,a_{22})=\sum_{i=1}^{22}a_i-59$ furthermore: $L=f(a_1,\dots,a_{22})-\lambda g(a_1,\dots,a_{22})$ $$\frac{\partial L}{\partial a_1}=-\frac{1}{(a_1+1)^2}-\lambda=0$$$$\vdots$$$$\frac{\partial L}{\partial a_{22}}=-\frac{1}{(a_{22}+1)^2}-\lambda=0$$This furthermore forces $a_1=\dots=a_{22}$ thus $a_1=\dots a_{22}=2.68...$, however by the conditions they must be integers, thus trivially minimum is only achieved only when they are $3$s and $2$s, however since their sum is $59$ and they are $22$ numbers, for minimum to be achieved there have to be $15$ threes and $7$ twos. Thus $f(2,2,\dots,3)=\frac{15}{4}+\frac{7}{3}>6$. QED And we are done!

F10tothepowerof34 wrote: The inequality is equal to: $\sum_{i=1}^{22}\frac{1}{a_i+1}>6$ This furthermore forces $a_1=\dots=a_{22}$ thus $a_1=\dots a_{22}=2.68...$, however by the conditions they must be integers, thus trivially minimum is only achieved only when they are $3$s and $2$s Why is this "trivially" true? You probably need some convexity argument to justify this (but then you could also solve the whole problem by a convexity argument instead of this Lagrange multipliers thing where, by the way, you forget to set and check the boundary conditions).

Tintarn wrote: Let $a_1,a_2,\dots,a_{22}$ be positive integers with sum $59$. Prove the inequality \[\frac{a_1}{a_1+1}+\frac{a_2}{a_2+1}+\dots+\frac{a_{22}}{a_{22}+1}<16.\] $\frac{a_1}{a_1+1}+\frac{a_2}{a_2+1}+\dots+\frac{a_{22}}{a_{22}+1}=T$ $22-T$$=\frac{1}{a_1+1}+\frac{1}{a_2+1}+\dots+\frac{1}{a_{22}+1}$ $\ge$$\frac{22^2}{(a_1+a_2+\dots+a_{22})+22} =\frac{22^2}{81} >5$ $\rightarrow$ $T<16$

Mathlover_1 wrote: Tintarn wrote: Let $a_1,a_2,\dots,a_{22}$ be positive integers with sum $59$. Prove the inequality \[\frac{a_1}{a_1+1}+\frac{a_2}{a_2+1}+\dots+\frac{a_{22}}{a_{22}+1}<16.\] $\frac{a_1}{a_1+1}+\frac{a_2}{a_2+1}+\dots+\frac{a_{22}}{a_{22}+1}=T$ $22-T$$=\frac{1}{a_1+1}+\frac{1}{a_2+1}+\dots+\frac{1}{a_{22}+1}$ $\ge$$\frac{22^2}{(a_1+a_2+\dots+a_{22})+22} =\frac{22^2}{81} >5$ $\rightarrow$ $T<16$ From your reasoning I got $T<17$

MatteD wrote: F10tothepowerof34 wrote: The inequality is equal to: $\sum_{i=1}^{22}\frac{1}{a_i+1}>6$ This furthermore forces $a_1=\dots=a_{22}$ thus $a_1=\dots a_{22}=2.68...$, however by the conditions they must be integers, thus trivially minimum is only achieved only when they are $3$s and $2$s Why is this "trivially" true? You probably need some convexity argument to justify this (but then you could also solve the whole problem by a convexity argument instead of this Lagrange multipliers thing where, by the way, you forget to set and check the boundary conditions). Convexity boils down to the same conclusions, you would still need some sort of minimum, the justification is immediate from Lagrange. The boundaries are already set from the problem and the equality constraint.

F10tothepowerof34 wrote: the justification is immediate from Lagrange How would your method work on this similar problem? Problem wrote: Let $a_1,\dots a_{15}$ be positive intergers with sum $50$. Find the minimum of $\sum_{i=1}^{15} \frac{1}{2a_i-7}$.

MatteD wrote: F10tothepowerof34 wrote: the justification is immediate from Lagrange How would your method work on this similar problem? Problem wrote: Let $a_1,\dots a_{15}$ be positive intergers with sum $50$. Find the minimum of $\sum_{i=1}^{15} \frac{1}{2a_i-7}$. $L=f(a_1\dots a_{15})-\lambda g(a_1\dots a_{15})=\sum_{i=1}^{15} \frac{1}{2a_i-7}-\lambda(a_1+\dots a_{15} -50)$ After the partial derivatives we get: $\frac{2}{2a_1-7}=\dots=\frac{2}{2a_{15}-7}\Longrightarrow a_1=\dots=a_{15}=3.333\dots$, same as above, for minimum they must be threes and fours. $LHS\ge \frac{10}{6-7}+\frac{5}{8-7}=-5$, thus the minimum is $-5$. Same principal as above.[\s]

F10tothepowerof34 wrote: $L=f(a_1\dots a_{15})-\lambda g(a_1\dots a_{15})=\sum_{i=1}^{15} \frac{1}{2a_i-7}-\lambda(a_1+\dots a_{15} -50)$ After the partial derivatives we get: $\frac{2}{2a_1-7}=\dots=\frac{2}{2a_{15}-7}\Longrightarrow a_1=\dots=a_{15}=3.333\dots$, same as above, for minimum they must be threes and fours. $LHS\ge \frac{10}{6-7}+\frac{5}{8-7}=-5$, thus the minimum is $-5$. Same principal as above. That is not even close. I honestly don't know what the minimum is, but simply taking $a_1 = \cdots = a_{14} = 3$ and $a_{15} = 8$ gives $-125/9 < -5$. $-5$ is not the maximum either, which you can see by taking $a_1 = \cdots = a_{14} = 1$ and $a_{15} = 36$.

Yes you are correct(made a dumb mistake). If we expand to positive reals minimum is achieved when you consider $a_1=\dots =a_{15}=3.333…$ After inserting you get that the minimum is equal to $-44.99999991$. ( Probably similar to integers)

Tintarn wrote: Let $a_1,a_2,\dots,a_{22}$ be positive integers with sum $59$. Prove the inequality \[\frac{a_1}{a_1+1}+\frac{a_2}{a_2+1}+\dots+\frac{a_{22}}{a_{22}+1}<16.\] $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ They ask to prove that: $$\sum_{i=1}^{22} \frac{a_i}{a_i+1}<16$$$$\Leftrightarrow 22-\sum_{i=1}^{22} \frac{1}{a_i+1}<16$$It is enough to prove that: $$\Leftrightarrow \sum_{i=1}^{22} \frac{1}{a_i+1}>6...(\alpha)$$$\text{Proof:}$ $\text{WLOG }a_1\le a_2\le a_3\le \ldots \le a_{21}\le a_{22}$ $$\sum_{i=1}^{22} a_i=59$$By the Pigeonhole principle: $$a_1\le \frac{59}{22}$$$$\Rightarrow a_1\le 2$$$\color{red} \boxed{\textbf{If }a_1=1:}$ $\color{red}\rule{24cm}{0.3pt}$ $$\Rightarrow \sum_{i=1}^{22} \frac{1}{a_i+1}=\frac{1}{2}+\sum_{i=2}^{22} \frac{1}{a_i+1}...(I)$$By Titu's Lemma: $$\sum_{i=2}^{22} \frac{1}{a_i+1}\ge \frac{21^2}{\sum_{i=2}^{22} (a_i+1) }$$$$\Rightarrow \sum_{i=2}^{22} \frac{1}{a_i+1}\ge \frac{441}{79}$$$$\Rightarrow \frac{1}{2}+\sum_{i=2}^{22} \frac{1}{a_i+1}\ge \frac{441}{79}+\frac{1}{2}>6$$By $(I):$ $$\Rightarrow \sum_{i=1}^{22} \frac{1}{a_i+1}>6_\blacksquare$$$\color{red}\rule{24cm}{0.3pt}$ $\color{red} \boxed{\textbf{If }a_1=2:}$ $\color{red}\rule{24cm}{0.3pt}$ $$\Rightarrow \sum_{i=2}^{22} a_i=57$$By the Pigeonhole principle: $$\Rightarrow a_2\le \frac{57}{21}$$$$\Rightarrow a_2\le 2$$Like $a_1\le a_2$ $$\Rightarrow a_2=2$$$$\Rightarrow \sum_{i=3}^{22} a_i=55$$By the Pingeonhole principle: $$\Rightarrow a_3\le \frac{55}{20}$$$$\Rightarrow a_3\le 2$$Like $a_2 \le a_3$ $$\Rightarrow a_3=2$$$$\Rightarrow \sum_{i=4}^{22} a_i=53$$By the Pingeonhole principle: $$a_4\le \frac{53}{19}$$$$\Rightarrow a_4\le 2$$Like $a_3\le a_4$ $$\Rightarrow a_4=2$$$$\Rightarrow \sum_{i=1}^{22} \frac{1}{a_i+1}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\sum_{i=5}^{22} \frac{1}{a_i+1}...(II)$$By Titu's Lemma: $$\sum_{i=5}^{22} \frac{1}{a_i+1}\ge \frac{18^2}{\sum_{i=5}^{22} (a_i+1)}$$$$\Rightarrow \sum_{i=5}^{22} \frac{1}{a_i+1}\ge \frac{324}{69}$$$$\Rightarrow \frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\sum_{i=5}^{22} \frac{1}{a_i+1}\ge \frac{324}{69}+\frac{4}{3}$$By $(II):$ $$\Rightarrow \sum_{i=1}^{22} \frac{1}{a_i+1}\ge \frac{324}{69}+\frac{4}{3} >6 _\blacksquare$$$\color{red}\rule{24cm}{0.3pt}$ $\color{red} \boxed{\textbf{Conclusion:}}$ $\color{red}\rule{24cm}{0.3pt}$ $$\sum_{i=1}^{22} \frac{a_i}{a_i+1}<16$$$\color{red}\rule{24cm}{0.3pt}$ $\color{blue}\rule{24cm}{0.3pt}$