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

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.

• Try setting $a+b+c \leq 1$ – basket Oct 8 '16 at 1:53
• Given non-negative $a,b,c,d$, with $a + b + c + d = 1$, we can set $d = 1$ and $a = b = c = 0$. In this case $ab + bc + ca = 0 < \frac{1}{4}$, which fails the inequality given. – Larry B. Oct 8 '16 at 2:04
• This isn't true. Let d=.9999925. a=b=c=.0000025. Not true. – fleablood Oct 8 '16 at 2:39
• @sufaid, was the intended problem about $ab+bc+cd$ or $ab+bc+cd+da$? The second is more natural and gives a stronger inequality? – zyx Oct 16 '16 at 16:51
• @zyx the problem is "prove $ab+bc+cd$ .I got my solution .I stuck on the first line which Michael Rozenberg do. – Sufaid Saleel Oct 17 '16 at 1:22

The reversed inequality with $ab+bc+cd$ is true because by AM-GM
$ab+bc+cd\leq ab+bc+cd+da=(a+c)(b+d)\leq\left(\frac{a+b+c+d}{2}\right)^2=\frac{1}{4}$.
• But there’s no $d$ in his desired inequality. – Lubin Oct 8 '16 at 3:36
• There is definitely something wrong with the original question since $ab+bc+ca\le\frac13(1-d)^2\le\frac13$. The OP has accepted your answer, so it seems your edit to the question was admissible, but changing a question to match your answer seems a bit dubious. – robjohn Oct 8 '16 at 7:58
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 $[0,\frac{1}{4}]$.