The answer is 652400, achieved when $x = 467$ and $y=z=466$. For simplicity's sake, let $\lfloor x \rfloor =a, \lfloor y \rfloor =b , \lfloor z \rfloor =a $, and $\alpha = x-a, \beta = y-b, \gamma = z-c$. Then the expression evaluates to \[ ab + bc + ca + a \beta + b \gamma + c \alpha. \]If $a + b + c = 1399$, then it is to apply a smoothness argument to show that our maximum is attained at that point. Explicitly, assume that $a \geq b + 2$. We claim taking $(a,b) \to (a-1, b+1)$ increases the maximum. Since each quantity is integral, it suffices to check that the following quantity is at least zero: \[ (a-1)(b+1) + c(a -1 + b+1) - ab - bc - ca = a -b -1 \geq 0 \]Now suppose for the sake of argument that $a + b + c = k < 1399$. By two uses of Cauchy-Schwarz, we have that \[ ab + bc + ca < \sqrt{(a^2 + b^2 + c^2)(b^2 + c^2 + a^2)} \leq \sqrt{\frac{ (a+b+c)^2 (b + c + a)^2}{9}} = \frac{k^2}{3} . \]Applying Cauchy-Schwarz again \[ a\beta + b \gamma + c \alpha \leq \sqrt{ {(a^2 + b^2 + c^2)(\beta^2 + \gamma^2 + \alpha^2)}} \leq \sqrt{\frac{(a+b+c)^2(\alpha + \beta + \gamma)^2}{9}} = \frac{2k}{3}, \]with the final inequality coming from $\alpha + \beta + \gamma \leq 2$ since $\alpha, \beta, \gamma <1$. Hence \[ \lfloor x \rfloor y + \lfloor y \rfloor z + \lfloor z \rfloor x \leq \frac{k^2 + 2k}{3}. \]The function $\frac{k^2 + 2k}{3}$ is obviously monotonic, so it suffices to check that this expression is less than 651934 when $k=1388$, which is indeed true.

Mr.C wrote: let $x,y,z$ be positive reals , such that $x+y+z=1399$ find the $$\max( [x]y + [y]z + [z]x ) $$( $[a]$ is the biggest integer not exceeding $a$) What is the relation of the title with the problem?

its a hint to use mixing

HKIS200543 wrote: The answer is 651934, achieved when $x = 467$ and $y=z=466$. Correct me if I'm wrong, but isn't the answer achieved by this example 652400? Since $x,y,z$ are integers, $\lfloor x \rfloor=x$ and then the answer would be $466^2+2*466*467=652400$

H.M-Deadline wrote: HKIS200543 wrote: The answer is 651934, achieved when $x = 467$ and $y=z=466$. Correct me if I'm wrong, but isn't the answer achieved by this example 652400? Since $x,y,z$ are integers, $\lfloor x \rfloor=x$ and then the answer would be $466^2+2*466*467=652400$ That would be correct. I've edited my solution.

Let $x=n+\alpha,\ y=m+\beta,\ z=r+\gamma$ such that $n,m,r\in \mathbb{Z}$ and $0\le \alpha,\beta,\gamma<1$. from the condition: $$n+m+r+\alpha+\beta+\gamma=1399$$we want to maximize $nm+mr+rn+n\beta+m\gamma+r\alpha$. Since $0\le \alpha+\beta+\gamma<3\Longrightarrow m+n+r\in\{1397,1398,1399\}$. we have the following three cases: case 1) $m+n+r=1397$: in this case $\alpha+\beta+\gamma=2$. suppose two numbers among $n,m,r$ differ at least $2$. WLOG $n-m\ge 2$. it's easy to see if we consider $(n-1,m+1,r)$ the sum $\lfloor x\rfloor y + \lfloor y\rfloor z + \lfloor z\rfloor x $ becomes $nm+mr+rn+n\beta+m\gamma+r\alpha+(n-m-1+\gamma-\beta)$. since $\gamma-\beta\ge -1$ and $n-m\ge 2\Longrightarrow n-m-1+\gamma-\beta\ge 0$ so we get a larger or equal sum. so we conclude that any two numbers in the set $\{n,m,r\}$ differ at most $1$. so maximum of $nm+mr+rn$ occurs when two of them are $466$ and the other $465$. WLOG $n=m=466,r=465$ so: $$nm+mr+rn+n\beta+m\gamma+r\alpha=650536+(932-\alpha)=651468-\alpha\le 651468$$equality occurs when $\alpha=0, n=m=466,r=465,\beta+\gamma=2$. case 2) $m+n+r=1398\Longrightarrow \alpha+\beta+\gamma=1$: Similarly the maximum of $mn+mr+rn$ is when $n=m=r=466$: $$nm+mr+rn+n\beta+m\gamma+r\alpha=651468+466\times 1=651934$$ case 3) $m+n+r=1399\Longrightarrow \alpha=\beta=\gamma=0$: Similarly the maximum of $mn+mr+rn$ is when $n=m=466,r=467$: $$nm+mr+rn+n\beta+m\gamma+r\alpha=652400$$ Hence combining three cases we conclude maximum occurs in the third case, $\boxed{\max( \lfloor x\rfloor y + \lfloor y\rfloor z + \lfloor z\rfloor x )=652400}$ and equality occurs when two numbers among $x,y,z$ are $466$ and the third is $467$.

Mr.C wrote: its a hint to use mixing Also the mixing solution alone is probably a $7^-$ if one doesn't prove the existence of maximum value of the function. In inequalities the existence of maximum is accepted without proof but this definitely needs that part. Having said that, the mixing solution can be completed by assuming that there is an example of a greater value than the one we have then using mixing concluding that if such an example exists there is also an integer-valued counterexample to the bound which can't happen. P.S: Maybe I am just wrong about how you want to use mixing.

matinyousefi wrote: Mr.C wrote: its a hint to use mixing Also the mixing solution alone is probably a $7^-$ if one doesn't prove the existence of maximum value of the function. In inequalities the existence of maximum is accepted without proof but this definitely needs that part. Having said that, the mixing solution can be completed by assuming that there is an example of a greater value than the one we have then using mixing concluding that if such an example exists there is also an integer-valued counterexample to the bound which can't happen. P.S: Maybe I am just wrong about how you want to use mixing. ok so for just being sure that i didnt messed up this is my solution . . . . well i'll just write my solution lets have $f(x,y,z)=[x]y+[y]z+[z]x, x+y+z=1399$ we want to find $\max f(x,y,z)$ in positive reals first we show that they "should" be integers , for the sake of contradiction lets write $x=x_1 +x_2 , y=y_1 + y_2 , z=z_1 + z_2$ (afcourse all of $z_2,x_2,y_2$ are not $0$) such that $z_1=[z],x_1=[x],y_1=[y]$ we will prove that if $z$ is the $ \min(x,y,z)$ then we have $f(x,y,z)<f(x_1,y_1,z_1+(z_2 + x_2 + y_2) )$ so lets write what we have $f(x,y,z)=x_1 y_1 + y_1 z_1 + x_1 z_1 + x_1 y_2 + y_1 z_2 + z_1 x_2$ on the other hand we have $f(x_1,y_1,z_1+(z_2 + x_2 + y_2 ))=x_1 y_1 + z_1 x_1 + z_1 y_1 + (x_2 + y_2 + z_2)(x_1 + y_1)$ and we have $x_1 y_2 + y_1 z_2 + z_1 x_2 <\max( x_1,y_1) (x_2 + y_2 +z_2)$ so for now we showed its maximum will be attaind for positive integers , and then again by using if $x-z>1$ then we have $f(x,y,z)<f(x-1,y,z+1)$ we will find the max of the function on $466,466,467$ and thats how my solution goes ... i guess its ok

Quote: Let $x$, $y$, $z$ be positive real numbers such that $x+y+z=1399$. Determine the maximum value of $[x]y + [y]z + [z]x$. We claim that the maximum value is $466^2+2\times 466\times 467$. Clearly, equality is possible (let $x=y=466$ and $z=467$). Therefore, it remains to prove that it is an upper bound for $[x]y + [y]z + [z]x$. \begin{align*} [x]y + [y]z + [z]x &= [x]y + [y]\left( (z-1)+1\right) + \left([z-1]+1\right) x\\ &= \left( [x]y + [y](z-1) + [z-1] x \right)+x+[y]\\ &\leq \frac{(x+y+z-1)^2}{3}+x+[y]\\ &=3\times 466^2 +x+[y]. \end{align*}Hence, if $x+[y]\leq 2\times 466$ (or similarly $y+[z]$ or $z+[x]$), the conclusion follows. Assume the contrary, i.e. $y+[z]$, $z+[x]$, $x+[y]> 2\times 466$. Thus, $$3\times 466 +1+[x]+[y]+[z]=x+y+z+[x]+[y]+[z]> 6\times 466$$and since $[x]\leq x$, we have $[x]+[y]+[z]\in \lbrace 3\times 466, 3\times 466 +1\rbrace$. If $[x]+[y]+[z]=3\times 466 +1$, all three of them are integers and we need to maximize $xy+yz+zx$ when $x+y+z=1399$. $$xy+yz+zx=\frac{1399^2-x^2-y^2-z^2}{2}$$which means we must minimize $x^2+y^2+z^2$, which occurs at $(466,466,467)$ and it's permutations (by MV or combinatorial form of Jensen's inequality). If $[x]+[y]+[z]=3\times 466$, then $\lbrace x\rbrace+\lbrace y\rbrace+\lbrace z\rbrace=1$. Therefore, one of these three (e.g. $\lbrace x\rbrace$) is not less than $\frac{1}{3}$. Hence, $$[x]y + [y]z + [z]x\leq \left(x-\frac{1}{3}\right) y +yz+zx.$$If $y\geq 1$, we are done and for $y<1$, the problem is trivial. $\blacksquare$

One of my friend has a solution but he doesnt have an Aops acc so all post it for you guys , step 1 lets just induct the whole thing for $x+y+z=3k+1$ and we want the $\max [x]y+[y]z+[z]x$ for $k=0$ that number is 0 (its quite trivial) lets see what happens lets say$\min(x,y,z)>1$ and $x+y+z=3k+4$ and $x=a+1,y=b+1,z=c+1$ so we want to max the $([a]+1)(b+1)+([b]+1)(c+1)+([c]+1)(a+1)$ which is $[a]b+[b]c+[c]a + (a+b+c)+3+([a]+[b]+[c])$ so lets have a claim, for all $k$ we claim the $\max$ is achieved by taking $(x,y,z)=(k,k,k+1)$ and the max is equal to $ 3k^2+2k$ now we have $a+b+c=3k+1$ so we have the bound of $3k^2+2k+3k+1+3+3k+1=3(k+1)^2+2(k+1)$ so its done now assume $\min(x,y,z)=x<1$ so we just want $\max [y]z+[z]x < 3(k+1)^2+2(k+1)$ for $x<1$ and $x+y+z=3k+4$ we have $[y]z+[z]x<z(x+y)=z(3k+4-z)<(3/2 k +2)^2<3(k+1)^2+2(k+1)$ for $k>0$ so the calim is proven and we have for $k=466$ the max is equal to $466(1400)$and we are done

we will solve the problem using following lemmas.

Lemma 1

Lemma 2

Lemma 3

Lemma 4

now we come back to problem. we know that $1 > \lbrace x\rbrace ,\lbrace y\rbrace,\lbrace z\rbrace \ge 0$ and $2 \ge \lbrace x\rbrace+\lbrace y\rbrace+\lbrace z\rbrace \ge 0 $ so we have three cases case 1: $ \lbrace x\rbrace+\lbrace y\rbrace+\lbrace z\rbrace = 0$ then $ x, y ,z \in \mathbb N$ and by lemma 1 maximum achieves when $x=y=466 , z=467$ case 2:$ \lbrace x\rbrace+\lbrace y\rbrace+\lbrace z\rbrace = 1$ WLOG assume$ x \ge y \ge z $then obviously $xy+yz+zt + x \ge [x]y + [y]z + [z]x $ and $3*466^2 +2*466 \ge \frac{(x+y+z)^2}{3}+x \ge xy + yz +zx +x $ case 3:$ \lbrace x\rbrace+\lbrace y\rbrace+\lbrace z\rbrace = 2$ WLOG assume$ x \ge y \ge z $then obviously $xy+yz+zt + 2x \ge [x]y + [y]z + [z]x $ and $3*466^2 +2*466 \ge \frac{(x+y+z)^2}{3}+2x \ge xy + yz +zx +2x $ these three cases implies that $3*466^2 +2*466 \ge \lfloor x\rfloor y + \lfloor y\rfloor z + \lfloor z\rfloor x $ and equality achieves when $x=y=466 , z=467$ and other permutations .

Answer is $652400$ Lemma (1)

Click to reveal hidden text

Also we know $[x] \leq x$ so , $$ [x]y + [y]z + [z]x \leq xy+yz+zx \leq \frac{(x+y+z)^2}{3} = 652400+\frac{1}{3}$$ Suppose that at least one of the numbers $x,y$ and $z$ is not correct. Since $z + y + x$ is correct, the sum of the decimal components of $x, y$ and $z$ is at least 1, so the decimal component of one of them is at least $\frac{1}{3} $. The rest is simple

Answer is $466^2+2.466.467=652400$ with example $(x,y,z)=(467,466,466)$. Denote $x=k+a, \ y=l+b, \ z=m+c$ where $0\leq a,b,c<1$. WLOG $k\geq l,m$. \[ S=[x]y + [y]z + [z]x=kl+lm+mk+kb+lc+ma\] $i)$ If $a+b+c=0,$ then $a=b=c=0$. $S=f(k,l,m)=kl+lm+mk$ Let $f(k,l,m)=S_{max}$. If $k,l,m$ are odd, then $f(k,\frac{l+m}{2},\frac{l+m}{2})>f(k,l,m)$ hence we get a contradiction. Let $l,m$ be even. If $l\neq m,$ then $f(k,\frac{l+m}{2},\frac{l+m}{2})>f(k,l,m)$ would yield a contradiction again. So $l=m$. \[kl+lm+mk=2kl+l^2=l^2+2l(1399-2l)=l(2798-3l)\leq 466^2+2.466.467\]Since $g(x)=x(2798-x)$ increases between $[0,1399]$ and decreases between $[1399,2798]$. If $l\leq 466,$ then $\frac{g(3l)}{3}\leq \frac{g(1398)}{3}=466.1400=466^2+2.466.467$ and if $l\geq 467,$ then $\frac{g(3l)}{3}\leq \frac{g(1401)}{3}=467.1397<466.1400$ $ii)$ If $a+b+c=1,$ then $k+l+m=1398$. \[S=kl+lm+mk+k(1-a-c)+lc+ma=kl+lm+mk+k+(m-k)a+(l-k)c\leq kl+lm+mk+k\]\[S\leq k(l+m+1)+lm\leq k(1399-k)+\frac{(1398-k)^2}{4}\]Let $932\geq 1398-k=t$. \[S\leq \frac{t^2}{4}+(t+1)(1398-t)=1398+t(1397-\frac{3t}{4})\overset{?}{\leq}466.1400\]$f(x)=\frac{3}{4}t(1397-\frac{3}{4}t)$ We can see that $f$ increases between $[0,931+\frac{1}{3}]$ and decreases between $[931+\frac{1}{3},932]$. Hence \[S_{max}\leq 1398+\frac{4}{3}f(t)\leq 1398+\frac{4}{3}.\max\{f(931),f(932)\}<466.1400\] $iii)$ If $a+b+c=2$, then $k+l+m=1397$. \[S=kl+lm+mk+2k-ka-kc+lc+ma\leq kl+lm+mk+2k=k(l+m+2)+lm\]\[S\leq k(1399-k)+lm\leq k(1399-k)+\frac{(1397-k)^2}{4}<k(1399-k)+\frac{(1398-k)^2}{4}\]Where $k\geq 466$. We can see that $S\leq 466.1400$ with the same method applied in second case as desired.$\blacksquare$