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I need to solve the next equation system:

Find all real numbers $a,b,c,d$ such that:

$$ \left\{ \begin{array}{c} a+b+c+d=20 \\ ab+ac+ad+bc+bd+cd=150 \\ \end{array} \right. $$

I tried something like this:

$b+c+d=20-a$

And i put the second equation like this

$a(b+c+d) + bc+bd+cd=150$

Getting

$20a-a^2 +bc+bd+cd=150$

But i see that this is useless, so i don't know how to start this problem.

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  • $\begingroup$ Hint: Compute $(a+b+c+d)^2$ $\endgroup$ – cansomeonehelpmeout Jan 29 '18 at 20:13
  • $\begingroup$ are These numbers supposed to be positive? $\endgroup$ – Dr. Sonnhard Graubner Jan 29 '18 at 20:14
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By $(a+b+c+d)^2=a^2+b^2+c^2+d^2+2(ab+ac+ad+bc+bd+cd)$, we find $$ a^2+b^2+c^2+d^2 = 100 $$

Also by Cauchy-Schwarz inequality, $(a+b+c+d)^2 \leq (1^2 + 1^2 + 1^2 +1^2)(a^2+b^2+c^2+d^2)$ and therefore we yields: $$ 400 \leq 400 $$

Then, the equality condition occurs if and only if $a=b=c=d=5$.

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$$0=3(a+b+c+d)^2-8(ab+ac+ad+bc+bd+cd)=$$ $$=(a-b)^2+(a-c)^2+(a-d)^2+(b-c)^2+(b-d)^2+(c-d)^2\geq0,$$ where the equality occurs for $a=b=c=d.$

Thus, $a=b=c=d=5.$

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