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Ab+bc+ac<=a^2+b^2+c^2

User Dudi Boy
by
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1 Answer

5 votes

Answer:

Now considering that

for all x :
x^(2) \geq 0

we show that


ab + bc + ac \leq a^(2) + b^(2) + c^(2)\\2ab + 2bc + 2ac \leq 2a^(2) + 2b^(2) + 2c^(2)\\\\0 \leq 2a^(2) + 2b^(2) + 2c^(2) -2ab -2bc -2ac \\\\\\0 \leq (a^(2) + b^(2) - 2ab) + (a^(2) + c^(2) - 2ac) + (b^(2) + c^(2) - 2bc) \\\\0 \leq (a+b)^2 + (a+c)^2 + (b+c)^2 \\\\

Explanation:

User Andrei Berenda
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