<s>[INST] How prove this $ab+bc+cd\le\dfrac{5}{4}$ let $a,b,c,d\in \Bbb R$ and $a,b,c,d>-1,a+b+c+d=0$
prove that
$$ab+bc+cd\le\dfrac{5}{4}$$
I have this solution
if $b\le c$, then
$$ab+bc+cd=a(b-c)-c^2\le -(b-c)-c^2=-(c-\dfrac{1}{2})^2+\dfrac{1}{4}-b\le\dfrac{1}{4}-b\le \dfrac{5}{4}$$
and then
$b>c$,as the same methods.I think this equality have other nice methods.Thank you
 [/INST] First we rewrite $ab+bc+cd= (a+c)(b+d) - ad = -(a+c)^2-ad$.
This expression can only be positive if either $a$ or $d$ is negative. Thus wlog we choose $d<0$ and $a>0$. Now we have $-(a+c)^2-ad <a-(a+c)^2$. If $a\leq1$, then this expression is $\leq1$ as well, thus wlog we assume $a>1$. In this case $a-(a+c)^2 < a-(a-1)^2 = -a^2+3a-1 = -\left(a-\frac{3}{2}\right)^2+\frac{5}{4}\leq \frac{5}{4}$.
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