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anemone

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- #1

anemone

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- #2

kaliprasad

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$a^2+5b^2+4c^2-4ab-4bc=0$

Or $a^2-4ab + 4b^2 + b^2 - 4bc + 4c^2 = 0$

Or $(a-2b)^2 + (b- 2c)^2= 0$

Because a,b,c are real so each part is zero

This gives the solution a = 2b and b = 4c

Or a = 4c, b =2c and so $a > b+ c$

so the answer is no

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