One of a,b,c,d is not greater than -1

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One of a,b,c,d is not greater than -1

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WakeUp

#1 h Nov 13, 2010, 2:03 PM • 2 Y

Y by Adventure10, Mango247

Let $a,b,c,d$ be real numbers such that
$a+b+c+d=-2$
$ab+ac+ad+bc+bd+cd=0$
Prove that at least one of the numbers $a,b,c,d$ is not greater than $-1$ .

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#2 h Nov 13, 2010, 10:24 PM • 1 Y

Y by Adventure10

WakeUp wrote:

Let $a,b,c,d$ be real numbers such that
$a+b+c+d=-2$
$ab+ac+ad+bc+bd+cd=0$
Prove that at least one of the numbers $a,b,c,d$ is not greater than $-1$ .

We have $4=(a+b+c+d)^2=a^2+b^2+c^2+d^2$ . Wlog we can assume $|a|\ge 1$ .
Suppose $a$ is greater than $-1$ , thus $a>1$ . But then $b+c+d< -3$ , so one of $b,c,d$ is smaller than $-1$ .

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#3 h Mar 2, 2011, 12:59 PM • 1 Y

Y by Adventure10

Put A = a + 1, B = b + 1, C = c + 1, D = d + 1.
In the first hypothesis, replace a by A - 1 etc. You find
A + B + C + D = 2.
Work similarly with the second hypothesis and use the result A + B + C + D = 2.
You find AB + AC + AD + BC + BD + CD = 0, thus A, B, C and D connot be all positive, which is the thesis.

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#4 h Nov 13, 2011, 5:32 AM • 2 Y

Y by Adventure10, Mango247

let us supppose $a,b,c,d>-1$
from $\sum ab=0$ we get $\sum a^2=4$
since $x^2$ is convex, $\sum a^2$ attains its maximum $4$ when $(a,b,c,d)=(-1,-1,-1,1)$
proved.

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#7 h Feb 1, 2025, 2:23 AM • 2 Y

Y by Nonononojeusson, JH_K2IMO

say that $a, b, c, d$ is all greater than $-1$
at least one of them should be under zero because $ab+ac+ad+bc+bd+cd=0$
also, if more than 3 is under zero, $ab+ac+ad+bc+bd+cd=0$ can't be established
so: wlog. $-1<a,b<0$
$a+b>-2$
the rest is all greater than zero, so $a+b+c+d$ is always greater than -2
so at least one should be below -1.

This post has been edited 2 times. Last edited by cmjun1109, Feb 1, 2025, 8:54 AM

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#8 h Apr 4, 2025, 10:56 AM

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I have more natural solution.
Let's assumme $a=x-1, b=y-1, c=z-1, d=w-1$ for $x,y,z,w>0$ . Then we have $x+y+z+w=2$ and $xy+xz+xw+yz+yw+wz=0$ , which is impossible.

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