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 $-1$.

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.

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.

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.