Prove that $ab+bc+cd \le \frac{1}{4} $ if $a+b+c+d =1$

Question

If $a,b,c,d$ are four nonnegative real numbers and $a+b+c+d=1$ then prove that $$ab+bc+cd \le \frac {1}{4} $$ It is the problem. I tried A.M.-G.M. Inequality, Cauchy-Swartz inequality. But I can't proceed. Somebody help me.

Answer 1

The reversed inequality with $ab+bc+cd$ is true because by AM-GM

$ab+bc+cd\leq ab+bc+cd+da=(a+c)(b+d)\leq\left(\frac{a+b+c+d}{2}\right)^2=\frac{1}{4}$.

Answer 2

The inequality is false.

The correct bounds are $0 \leq ab+bc+ca \leq \frac{1}{3}$ with the minimum when $ab=bc=ca=0$ (such as $a=b=c=0$) and the maximum when $a=b=c=\frac{1}{3}$.

If that is taken as evidence that the problem was supposed to be about $ab+bc+cd$ or $ab+bc+cd+da$, the other answer applies and the range is $[0,\frac{1}{4}]$.