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Let $x_1 , x_2 , x_3 , x_4 , x_5$ be non-negative real numbers such that $x_1 + x_2 + x_3 + x_4 + x_5=5$ . Determine the maximum value of $x_1x_2+x_2x_3+x_3x_4+x_4x_5$.

Normally in such questions I use the fact that the equation is symmetric and thus extremum is attained when all the variables are equal , but this can not be done here , and I have spent quite a long time on this but nothing worth-mentioning came to my mind . Could someone please help me find the maximum value ?

Thanks !

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4 Answers 4

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$x_1x_2+x_2x_3+x_3x_4+x_4x_5=(x_1+x_3+x_5)(x_2+x_4) - (x_2x_5+x_1x_4) $

Now we try to find maximum value of $(x_1+x_3+x_5)(x_2+x_4)$ when $(x_1+x_3+x_5)+(x_2+x_4)=5$ And try to minimize the value of $(x_2x_5+x_1x_4) $.

Take, $a=(x_1+x_3+x_5)$ and $b=(x_2+x_4)$ By , A.M. $\ge $ G.M. $\implies$ $\sqrt(ab) \le \frac{a+b}{2} \implies (ab) \le (\frac{5}{2})^2 $ So, max value of $(x_1+x_3+x_5)(x_2+x_4)$ is $(\frac{5}{2})^2 $

And , clearly, minimum value of $(x_2x_5+x_1x_4) $ is $0$. So , max value of $x_1x_2+x_2x_3+x_3x_4+x_4x_5$ is $(\frac{5}{2})^2 $

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Hint: Reduce the number of variables, $$x_1x_2+x_2x_3+x_3x_4+x_4x_5\leq x_2(x_1+x_3)+(x_1+x_3)x_4+x_4x_5\leq\dots$$

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Hint. Note that $(x_1+x_3+x_5)+(x_2+x_4)=5$ and $$ x_1x_2+x_2x_3+x_3x_4+x_4x_5\leq(x_1+x_3+x_5)(x_2+x_4). $$ Can you end now?

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  • $\begingroup$ Yes , I was able to end it from here , thanks $\endgroup$
    – ARROW
    Commented Aug 6, 2020 at 11:47
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In general, for $n\ge2$, given the constraint $$x_1+\cdots+x_n=c,\qquad x_i\ge0$$ then $$x_1x_2+\cdots+x_{n-1}x_n\le\frac{c^2}{4}$$ with the maximum achieved by $x_1=x_2=c/2$, $x_i=0$ ($i>2$) (or any other pair equal).

Using Lagrange multipliers, a critical point is achieved when $$x_2=\lambda,\quad x_{i-1}+x_{i+1}=\lambda,\quad x_{n-1}=\lambda$$ As $x_2=x_2+x_4$, then $x_4=0$. Since one of the variables has to be zero, we can without loss of generality let it be $x_n$, since it contributes to just one term in the expression. Then the problem reduces to the same one for $n-1$ terms, which achieves the maximum at $x_1=x_2=c/2$, by induction. The starting steps for $n=2$ and $n=3$ are trivial.

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  • $\begingroup$ Actually, I am getting there but , Langrange multipliers is not familiar to me yet , I'll comeback to this answer once I learn it , thanks $\endgroup$
    – ARROW
    Commented Aug 6, 2020 at 11:48

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