Tintarn wrote:

Sorry，but can you be more specific with the Lagrange multipliers?I didn't get the formula which you have got.(plus.If the Lagrange multipliers is allowed in the CGMO?)

I did not use Lagrange multipliers. I just said that one could also use Lagrange multipliers to arrive at the same equations.

Tintarn wrote: I did not use Lagrange multipliers. I just said that one could also use Lagrange multipliers to arrive at the same equations. So how can you get this formula with a,b,c? Without Lagrange multipliers,how can we get that by shifting?

Chronokeeper wrote: So how can you get this formula with a,b,c? Without Lagrange multipliers,how can we get that by shifting? It is the same argument as the one used before: Let's say we increase $a$ by $\varepsilon$ and decrease $b$ by $\varepsilon$ (possible since both are clearly strictly positive). Then we get the expression $(a+\varepsilon)^4(b-\varepsilon)^2(b+c-\varepsilon)^2$ (the rest is constant). For this to have a local maximum at $\varepsilon=0$ we need $\frac{4}{a}=\frac{2}{b}+\frac{2}{b+c}$ which is one of the two equations I gave. I am sure you can now figure out how to get the other one on your own.

Solution: It's easy to show that $x_3=x_5=x_7=x_9=x_{11}=0$when F achives maximum.(Othwise let $x_{2k}'=x_{2k}'+x_{2k-1}'$) For convinence, say $a=x_1,b=x_2,c=x_4,d=x_6,e=x_8,f=x_{10}$ Then $F(a,b,c,d,e,f)=(a+b)(b+c)(c+d)(d+e)(e+f)(f+a)acdef$ Fix multi-set $\{a,b,c,d,e,f\}=\{u \leq v \leq w \leq x \leq y \leq z\}$ It is easy to show that F achieves maximum when $b=u$ Notice that $a'=c'=\frac{a+c}{2},f'=d'=\frac{f+d}{2}$ makes $acdf$and$(b+a)(b+c)$,$(e+d)(e+f)$,$(a+f)(c+d)$ larger(or don't change when a=c d=f), so $a=c,d=f$ And by $F_{max}(a,b,c,d,e,f) \ge F(a,b,d,c,e,f)$We know $(e-b)(c-d)\ge 0$ By $F_{max}(a,b,c,d,e,f) \ge F(a,b,c,e,d,f)$We know $(f-c)(d-e)=(d-c)(d-e)\ge 0$ Then $e\ge d$ Q.E.D. (Easy to check the equality can't hold,since we have deduce F to a four varible function with 2a+b+2d+e=1,a simple calculation shows if c=d we can take d'=d+2t and c'=c-t for small t)

szjzc2018 wrote: And by $F_{max}(a,b,c,d,e,f) \ge F(a,b,d,c,e,f)$ we can know $e \ge d$(use $c \ge b$) Q.E.D. (Easy to check the equality can't hold) Uh, quite wrong unfortunately. 1. The condition $F(a,b,c,d,e,f) \ge F(a,b,d,c,e,f)$ is clearly equivalent to $(b+c)(d+e) \ge (b+d)(c+e)$ and hence to $(e-b)(c-d) \ge 0$. I don't see how $e \ge d$ follows from here (and how $c \ge b$ is relevant). Why can't we just have $c>d>e>b$, say? 2. Even if you get $e \ge d$, I don't see how it is "easy to see" that equality can't hold.

Tintarn wrote: szjzc2018 wrote: And by $F_{max}(a,b,c,d,e,f) \ge F(a,b,d,c,e,f)$ we can know $e \ge d$(use $c \ge b$) Q.E.D. (Easy to check the equality can't hold) Uh, quite wrong unfortunately. 1. The condition $F(a,b,c,d,e,f) \ge F(a,b,d,c,e,f)$ is clearly equivalent to $(b+c)(d+e) \ge (b+d)(c+e)$ and hence to $(e-b)(c-d) \ge 0$. I don't see how $e \ge d$ follows from here (and how $c \ge b$ is relevant). Why can't we just have $c>d>e>b$, say? 2. Even if you get $e \ge d$, I don't see how it is "easy to see" that equality can't hold. Sorry,I made some typo in my solution.Now it is OK.

Tintarn wrote: Chronokeeper wrote: So how can you get this formula with a,b,c? Without Lagrange multipliers,how can we get that by shifting? It is the same argument as the one used before: Let's say we increase $a$ by $\varepsilon$ and decrease $b$ by $\varepsilon$ (possible since both are clearly strictly positive). Then we get the expression $(a+\varepsilon)^4(b-\varepsilon)^2(b+c-\varepsilon)^2$ (the rest is constant). For this to have a local maximum at $\varepsilon=0$ we need $\frac{4}{a}=\frac{2}{b}+\frac{2}{b+c}$ which is one of the two equations I gave. I am sure you can now figure out how to get the other one on your own. Thank you!

Can someone please check this: if there are mistakes, please kindly point them out. Thank you! Note when $x_3$ appears, it always appears with $+x_4$ so we may assume $x_3=0$. Same thing for $x_5, x_7, x_9, x_{11}$. I am left with $(x_2+x_4)x_4 (x_4+x_6) x_6 (x_6+x_8) x_8 (x_8+x_{10}) x_{10} (x_{10}+x_1) x_1(x_1+x_2)$ Now invoke symmetry: I claim $x_1=x_4, x_6=x_{10}$ can be proven via amgm $x_1x_4$, $(x_1+x_2)(x_2+x_4)$, $(x_4+x_6)(x_{10}+x_1)$, $x_6x_{10}$, blah So I am left with $(x_2+x_1)^2 x_1^2 (x_1+x_6)^2 x_6^2 (x_6+x_8)^2 x_8$ Subject to $x_2+2x_1+2x_6+x_8=1$ By swapping $(x_2,x_8)$ and $(x_1, x_6)$ we can see $x_2<x_8$ and $x_6<x_1$ Rewrite this as $(2x_2+x_1)^2 x_1^2 (x_1+x_6)^2 x_6^2 (x_6+2x_8)^2 x_8$ subject to $x_2+x_1+x_6+x_8=\frac 12$ Changing $(x_2, x_1)$ to $(0, x_1+x_2)$ increases the product so $x_2=0$ So I am maximizing $a^4 (a+b)^2 b^2 (b+2c)^2 c$ subject to $a+b+c$ fixed. Wts: $b<c$ (here $a=x_1, b=x_6, c=x_8$) We bash this with Lagrange Multipliers. I am trying to maximize $a^4 (a+b)^2 b^2 (b+2c)^2 c$ subject to $a+b+c=\frac 12$. For laziness assume $a+b+c=1$. Clearly, $(a,b,c)$ lies in the compact region $[0,1]^3$. Let $f(a,b,c)=a^4 (a+b)^2 b^2 (b+2c)^2 c$ and $g(a,b,c)=a+b+c$. Then $\nabla f(a,b,c) = \langle \frac{df}{da}, \frac{df}{db}, \frac{df}{dc} \rangle = f\langle \frac{4}{a}+\frac{2}{a+b}, \frac{2}{a+b} + \frac{2}{b} + \frac{2}{b+2c}, \frac{4}{b+2c}+\frac 1c \rangle$ $\nabla g(a,b,c) = \langle 1,1,1\rangle$ Therefore, $\nabla f=t\nabla g\rightarrow \frac{4}{a}+\frac{2}{a+b}=\frac{2}{a+b} + \frac{2}{b} + \frac{2}{b+2c}=\frac{4}{b+2c}+\frac 1c$ $\frac{1}{a}= + \frac{1}{b} + \frac{1}{b+2c}$ and $\frac 1c + \frac{2}{b+2c} = \frac{2}{a+b} + \frac{2}{b} $ $\frac 1c + \frac 1a =\frac{2}{a+b}+\frac{3}{b}$ If $c<b$ then $LHS > \frac 1c > \frac 1b $. Since $a>b$, RHS is at most $\frac 1b (1/2+1/3)$

CANBANKAN wrote: Can someone please check this: if there are mistakes, please kindly point them out. Thank you! Sounds good! (I believe it is more or less isomorphic to my solution.)

CANBANKAN wrote: Can someone please check this: if there are mistakes, please kindly point them out. Thank you! Note when $x_3$ appears, it always appears with $+x_4$ so we may assume $x_3=0$. Same thing for $x_5, x_7, x_9, x_{11}$. I am left with $(x_2+x_4)x_4 (x_4+x_6) x_6 (x_6+x_8) x_8 (x_8+x_{10}) x_{10} (x_{10}+x_1) x_1(x_1+x_2)$ Now invoke symmetry: I claim $x_1=x_4, x_6=x_{10}$ can be proven via amgm $x_1x_4$, $(x_1+x_2)(x_2+x_4)$, $(x_4+x_6)(x_{10}+x_1)$, $x_6x_{10}$, blah So I am left with $(x_2+x_1)^2 x_1^2 (x_1+x_6)^2 x_6^2 (x_6+x_8)^2 x_8$ Subject to $x_2+2x_1+2x_6+x_8=1$ By swapping $(x_2,x_8)$ and $(x_1, x_6)$ we can see $x_2<x_8$ and $x_6<x_1$ Rewrite this as $(2x_2+x_1)^2 x_1^2 (x_1+x_6)^2 x_6^2 (x_6+2x_8)^2 x_8$ subject to $x_2+x_1+x_6+x_8=\frac 12$ Changing $(x_2, x_1)$ to $(0, x_1+x_2)$ increases the product so $x_2=0$ So I am maximizing $a^4 (a+b)^2 b^2 (b+2c)^2 c$ subject to $a+b+c$ fixed. Wts: $b<c$ (here $a=x_1, b=x_6, c=x_8$) We bash this with Lagrange Multipliers. I am trying to maximize $a^4 (a+b)^2 b^2 (b+2c)^2 c$ subject to $a+b+c=\frac 12$. For laziness assume $a+b+c=1$. Clearly, $(a,b,c)$ lies in the compact region $[0,1]^3$. Let $f(a,b,c)=a^4 (a+b)^2 b^2 (b+2c)^2 c$ and $g(a,b,c)=a+b+c$. Then $\nabla f(a,b,c) = \langle \frac{df}{da}, \frac{df}{db}, \frac{df}{dc} \rangle = f\langle \frac{4}{a}+\frac{2}{a+b}, \frac{2}{a+b} + \frac{2}{b} + \frac{2}{b+2c}, \frac{4}{b+2c}+\frac 1c \rangle$ $\nabla g(a,b,c) = \langle 1,1,1\rangle$ Therefore, $\nabla f=t\nabla g\rightarrow \frac{4}{a}+\frac{2}{a+b}=\frac{2}{a+b} + \frac{2}{b} + \frac{2}{b+2c}=\frac{4}{b+2c}+\frac 1c$ $\frac{1}{a}= + \frac{1}{b} + \frac{1}{b+2c}$ and $\frac 1c + \frac{2}{b+2c} = \frac{2}{a+b} + \frac{2}{b} $ $\frac 1c + \frac 1a =\frac{2}{a+b}+\frac{3}{b}$ If $c<b$ then $LHS > \frac 1c > \frac 1b $. Since $a>b$, RHS is at most $\frac 1b (1/2+1/3)$ I don't know whether my solution is right or not and I just take x_2=1-x_1-x_4-x_6-x_8-x_10 and I suppose the max is when x_6+x_8=P and this time $(1-x_4-x_6-x_8-x_10), $(1-x_1-x_6-x_8-x_10), x_1, x_4, x_10, $(x_6+x_8), $(x_10+x_1) are constant value and we only leave this one $(x_10+x_8)x_6x_8 and we can use AG to find out that x_8=\frac{2P+x_10}{3} and it's bigger than x_6

So you are fixing variables $x_1,x_2,x_4,x_{10}, x_6+x_8$ and moving $x_6, x_8$? I think you have $(x_4+x_6)x_6x_8(x_8+x_{10})$.... to show $x_6<x_8$ I think you need to show $x_4>x_{10}$ So yeah after you show that $x_2=0, x_1=x_4, x_6=x_{10}$ this becomes showing $a=x_1=x_4 > x_6=x_{10}=b$. But that is clear since $a(b+c)=b(b+2c)$. Your solution does work but it still doesn't avoid Lagrange Multipliers...

CANBANKAN wrote: So you are fixing variables $x_1,x_2,x_4,x_{10}, x_6+x_8$ and moving $x_6, x_8$? I think you have $(x_4+x_6)x_6x_8(x_8+x_{10})$.... to show $x_6<x_8$ I think you need to show $x_4>x_{10}$ So yeah after you show that $x_2=0, x_1=x_4, x_6=x_{10}$ this becomes showing $a=x_1=x_4 > x_6=x_{10}=b$. But that is clear since $a(b+c)=b(b+2c)$. Your solution does work but it still doesn't avoid Lagrange Multipliers... Yes that's my fault I forgot to say that and I maybe take that for granted

And I think that my first time using AG is wrong it should be changed into Lagrange Multipliers（but I absolutely don't know whether I can use it in CGMO

You can rephrase Lagrange Multipliers into something more elementary quite easily with derivatives (especially and in this case, you only have two variables obeying their sum is constant). Are derivatives not allowed in CGMO? That'd be surprising. If that fails, see Tintarn's shifting argument: Tintarn wrote: Then replacing $(x_1,x_2,x_4)$ by $(x_1-\varepsilon,x_2+2\varepsilon,x_4-\varepsilon)$ for some $\varepsilon$ sufficiently close to $0$ (not neccessarily positive) as a function of $\varepsilon$ the expression up to constants just becomes $(x_1-\varepsilon)(x_1+x_{10}-\varepsilon)(x_4-\varepsilon)(x_4+x_6-\varepsilon)(x_1+x_2+\varepsilon)(x_2+x_4+\varepsilon)$. For this to have a local maximum at $\varepsilon=0$ we see that we must have $\frac{1}{x_1}+\frac{1}{x_4}+\frac{1}{x_1+x_{10}}+\frac{1}{x_4+x_6}=\frac{1}{x_1+x_2}+\frac{1}{x_2+x_4}$

I believe only a small part of derivatives (the part taught in normal high-school classes) can be used freely in contests below CMO level, and that part does not include Lagrange Multipliers

Quote: For concreteness, let us note that \[F(x_1,\dots,x_{11})=(x_1+x_2)(x_2+x_3+x_4)(x_3+x_4)(x_4+x_5+x_6)(x_5+x_6)(x_6+x_7+x_8)(x_7+x_8)(x_8+x_9+x_{10})(x_9+x_{10})(x_{10}+x_{11}+x_1)(x_{11}+x_1).\]Note that $x_3$ only appears with $x_4$. Hence if we decrease $x_3$ and increase $x_4$ by the same amount, the expression grows. Hence, when the maximum occurs, we must have $x_3=0$ and similarly $x_5=x_7=x_9=x_{11}=0$. Hence our expression reduces to \[(x_1+x_2)(x_2+x_4)x_4(x_4+x_6)x_6(x_6+x_8)x_8(x_8+x_{10})x_{10}(x_{10}+x_1)x_1.\]Suppose that $x_2>0$. Then replacing $(x_1,x_2,x_4)$ by $(x_1-\varepsilon,x_2+2\varepsilon,x_4-\varepsilon)$ for some $\varepsilon$ sufficiently close to $0$ (not neccessarily positive) as a function of $\varepsilon$ the expression up to constants just becomes $(x_1-\varepsilon)(x_1+x_{10}-\varepsilon)(x_4-\varepsilon)(x_4+x_6-\varepsilon)(x_1+x_2+\varepsilon)(x_2+x_4+\varepsilon)$. For this to have a local maximum at $\varepsilon=0$ we see that we must have $\frac{1}{x_1}+\frac{1}{x_4}+\frac{1}{x_1+x_{10}}+\frac{1}{x_4+x_6}=\frac{1}{x_1+x_2}+\frac{1}{x_2+x_4}$ which is absurd. Hence we must indeed have $x_2=0$. Now we are left with the expression \[x_4^2(x_4+x_6)x_6(x_6+x_8)x_8(x_8+x_{10})x_{10}(x_{10}+x_1)x_1^2.\]Next, we note that by moving $x_4$ and $x_1$ as well as $x_6$ and $x_{10}$ to their respective mean, we just increase the product of opposite factors and hence the whole expression. We may henceforth assume $x_1=x_4=a, x_6=x_{10}=b, x_8=c$ so that $2a+2b+c=1$, our expression becomes \[a^4(a+b)^2b^2(b+c)^2c\]and we need to show $c>b$. Repeating the shifting argument with the remaining variables (which are all non-zero) or, equivalently, applying Lagrange multipliers, we get the equations \[\frac{2}{a}+\frac{1}{a+b}=\frac{1}{a+b}+\frac{1}{b}+\frac{1}{b+c}=\frac{2}{b+c}+\frac{1}{c}.\]W.l.o.g. put $c=1$. Solving the first equation for $a$ and inserting into the second one, we get $b(b+1)(4b+6)=(2b+1)(3b+4)$ and hence $4b^3+4b^2-5b-4=0$. But clearly this equation has a unique positive root $b>1$. Done. My students say there are two typos at the end. Should it be $4b^3+5b^2-4b-4=0$ and $b<1$?

Here is a non-calculus solution. Throughout the solution whenever we use $\epsilon$, we are referring to a very small positive number. First, let's write out the $y_i$s. Note that \begin{align*} y_1 &= x_1+x_2 & y_5 &= x_5+x_6 & y_9 &= x_9+x_{10} \\ y_2 &= x_2+x_3+x_4 & y_6&=x_6+x_7+x_8 & y_{10}&=x_{10}+x_{11}+x_1 \\ y_3 &= x_3+x_4 & y_7 &= x_7+x_8 & y_{11} &= x_{11}+x_1 \\ y_4 &= x_4+x_5 & y_8 &= x_8 + x_9 + x_{10} \end{align*}Observe that the $x_3$s only appears with an $x_4$. This implies that swapping $(x_3, x_4)$ with $(0, x_3+x_4)$ will increase the product $F$ unless $x_3=0$. Similarly we must have $x_5=x_7=x_9=x_{11}=0$. Now suppose that $x_2 >0$. Notice that if $x_1 \geq x_4$, swapping $(x_4, x_2)$ with $(x_4 + \epsilon, x_2- \epsilon)$ increases $F$ since $x_4(x_1+x_2)<(x_4 + \epsilon)(x+1+x_2 - \epsilon)$ and otherwise, swapping $(x_1, x_2)$ with $(x_1 + \epsilon, x_2 - \epsilon)$ does so as well. Hence, we must have $x_2=0$. Hence our expression becomes \[F=x_1^2x_4^2x_6x_8x_{10}(x_1+x_{10})(x_{10}+x_8)(x_8+x_6)(x_6+x_4)\]We show that $x_1 = x_4$ and $x_6 = x_{10}$. We may assume that $x_1+x_{10} \geq x_6+x_4$. If $x_1>x_4$ it is not difficult to see that simply swapping $(x_1, x_4)$ with $(x_1- \epsilon, x_4 + \epsilon)$ increases our product. On the other hand, if $x_1 < x_4$ we must have $x_{10} > x_6$ in which case we can swap $(x_{10}, x_6)$ with $(x_{10} - \epsilon, x_6 + \epsilon)$ to increase our product. Therefore, we must have $x_1=x_4$. We can similarly show that $x_6 = x_{10}$. For the sake of brevity, let $x_1=a$, $x_6=b$, and $x_8=c$. Then our expression is \[a^4b^2c(a+b)^2(b+c)^2\]with the condition that $2a+2b+c=1$. Suppose for contradiction that $b \geq c$. If $a<b$, swapping $(a, b)$ with $(a + \epsilon, b - \epsilon)$ increases our expression since $a^2b^2 < (a + \epsilon)^2(b- \epsilon)^2$ and $a^2(b+c)^2 < (a+\epsilon)^2(b+c-\epsilon)^2$. If $a>c$, swapping $(b, c)$ with $(b- \epsilon, c + 2\epsilon)$ also increases our expession because \[b^2c \approx (b^2-2b\epsilon + \epsilon^2)(c+2\epsilon)=(b-\epsilon)^2(c+2\epsilon) \textrm{ and}\]\[ a^2(b+c)^2 < (a+\epsilon)^2(b+c-\epsilon)^2\]Hence, we must have $a \leq c \leq b$ and $b \leq a$. This implies that $a=b=c = \frac{1}{5}$ in which case it is easy to show that $F$ doesn't take its maximum value by shifting.