If $a, b, c, d, e$ and $f$ are non negative real numbers such that $a + b + c + d + e + f = 1$, then what is the maximum value of $ab + bc + cd + de + ef$?
[Math] the Maximum Value
inequality
Related Solutions
$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 $
Because$:$ $$\dfrac{a}{18} + \dfrac{1}{9} - \dfrac{1}{a^2-4a+9} = \dfrac{a(a-1)^2}{18(a^2-4a+9)} \geq 0,$$ $$\therefore \dfrac{1}{a^2-4a+9} \leq \dfrac{a}{18} + \dfrac{1}{9}.$$ So$:$ $$ \sum \dfrac{1}{a^2-4a+9} \leq \dfrac{a+b+c}{18} + \dfrac{1}{3} = \dfrac{7}{18}.$$ Equality occur when $a:b:c=1:0:0$ or any permution.
Update. Let $$f(a) = (\,ma + n\,)(\,a^2-4a+9\,) - 1.$$ We will try to find $m,\,n$ such that $f(a) \geqslant 0$ for all $a \in [\,0,1\,].$
Proof 1. Let $a = 0$ and $a =1$ we get $$\{9n = 1,\; 6(m+n) = 1\}.$$ Solve equation we get $m = \dfrac 1 {18},$ $n = \dfrac 1 9.$
Proof 2. Write inequality as $$f(a) = a \Big[\,a^2m-(4m-n)a+9m-4n\,\Big] +9n-1.$$ It's easy choose $n = \dfrac{1}{9},$ we get $$f(a) = a \Big[\,ma^2-(4m-n)a+9m-4n\,\Big].$$ For a quadratic polynomial $ma^2-(4m-n)a+9m-4n,$ we have$:$ $$\Delta = -(10m+n)(2m-n) = -\left(10m+\dfrac 1 9\right)\left(2m-\dfrac 1 9\right).$$ We need $m\geqslant 0.$ So choose $2m = \dfrac{1}{9}$ which means $m = \dfrac{1}{18}.$
Best Answer
Note that lyj's comment pretty much answers this question. On top of that, we can achieve this maximum by taking $(a, b, c, d, e, f) = (0, 0, 11/32, 1/2, 5/32, 0)$.
I.e., We finish showing the following.
The quantity in question has an upper bound $1/4$ (by lyj's argument).
The upper bound $1/4$ can be achieved.