2

Non-negative real numbers $a,b,c,d$ are such that $a+b+c+d=2$. Prove or disprove that $$ab+bc+cd+da\leq1$$ I see there are multiple equality cases, where $(a,b,c,d)$ is for example $(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2})$, $(\frac{3}{4},\frac{1}{2},\frac{1}{4},\frac{1}{2})$, $(\frac{3}{4},\frac{5}{8},\frac{1}{4},\frac{3}{8})$.

I suspect it's true, and maybe it can be proven with rearrangement, but I have not found a way.

It's reminiscent of Chebyshev's inequality, since the desired inequality is equivalent to $$4(ab+bc+cd+da)\leq(a+b+c+d)(b+c+d+a)$$ But we cannot assume that $(a,b,c,d)$ and $(b,c,d,a)$ are oppositely ordered.

Ashton Parks
  • 105
  • 3
    Hint: make a substitution of $u = a + c$ and $v = b+d$. – Theo Bendit Oct 21 '22 at 01:39
  • Thank you, Theo. If $u=a+c$ and $v=b+d$, then the desired inequality is equivalent to $(u-1)^2\geq 0$. Now I'll investigate whether this holds for more variables, i.e. $\sum_{i=1}^{n-1} a_i a_{i+1} + a_n a_1\leq\frac{4}{n}$. – Ashton Parks Oct 21 '22 at 01:56
  • Generalisation here (https://math.stackexchange.com/questions/4559447), for future reference. – Ashton Parks Oct 23 '22 at 22:54

2 Answers2

2

Let $x = a+c$ and $y=b+d$. Then, we have $xy = (a+c)(b+d) = ab + ad + bc + cd$ which is the quantity we want to bound. So, we want to show that if $x+y = 2$, then $xy \leq 1$.

We just use basic algebra now: $x=2-y$, so $xy = y(2-y)$. Now you can rearrange, $y(2-y) = -(y-1)^2 +1$. Since $(y-1)^2 \geq 0$, it follows that $-(y-1)^2\leq 0$ and so $xy = -(y-1)^2+1 \leq 1$.

You could also just differentiate $-y^2 + 2y$, to see that you have a local maximum when $y=1$.

Fred T
  • 664
  • Thanks Fred. Do you know if this can be generalised? In other words, $$\sum_{i=1}^{n-1}(a_i a_{i+1})+a_n a_1\leq\frac{4}{n}$$ – Ashton Parks Oct 21 '22 at 02:36
  • @AshtonParks I doubt this method will work. This relies on the factorisation of $ab + bc + cd + da$ as $(a+c)(b+d)$, which doesn't generalise. – Fred T Oct 22 '22 at 07:30
2

We have: $ab+bc+cd+da = (a+c)(b+d) \le \dfrac{((a+c)+(b+d))^2}{4}= \dfrac{(a+b+c+d)^2}{4}= 1$.

Wang YeFei
  • 6,390