$a,b,c,d$ are real numbers summing to zero. Let $M=ab+bc+cd$ and $N=ac+ad+bd$. Prove that at least one of the sums $20M+17N$ and $20N+17M$ is non-positive.
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Parcly Taxel
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Aytajdadashova
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1Where did you come across this question, please? – Gerry Myerson May 10 '17 at 07:27
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What you have tried? – Harsh Kumar May 10 '17 at 07:33
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1I don't know my teacher gives to us – Aytajdadashova May 10 '17 at 07:35
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1I try to simplfy but i can't – Aytajdadashova May 10 '17 at 07:35
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Looks like it is worth starting with the fact that $(a+b+c+d)^2=0$ ... – Paul Aljabar May 10 '17 at 07:52
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Your teacher in what class, please? Is it homework? Is it an exam? – Gerry Myerson May 10 '17 at 13:04
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differential equation ,not homework but likely that – Aytajdadashova May 10 '17 at 14:54
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I'd really like to see how a question like this could come up in the study of differential equations! It would make your question so much more interesting, if you could include that. – Gerry Myerson May 11 '17 at 04:58
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You have that $ (a+b+c+d)^{2}=a^{2}+b^{2}+c^{2}+d^{2}+2(M+N) $ so since $ a+b+c+d=0 $, you get that $ 2(M+N)=-(a^{2}+b^{2}+c^{2}+d^{2}) \leq 0 $.
Now let $ S_{1}=20M+17N $ and $ S_{2}=20N+17M $. Then $ S_{1}+S_{2}=37(M+N) \leq 0 $ by the above so at least one of them must be non-positive, otherwise their sum would be greater than $ 0 $ , which is clearly absurd by the above.
Andrei
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