2003 Indonesia MO Problems/Problem 5

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Problem

For every real number $a,b,c$, prove the following inequality

\[5a^2 + 5b^2 + 5c^2 \ge 4ab + 4bc + 4ca\]

and determine when the equality holds.

Solution

By the Trivial Inequality, $(a-2b)^2 \ge 0,$ $(b-2c)^2 \ge 0,$ and $(c-2a)^2 \ge 0.$ Adding all the equations, expanding the terms, and rearranging the resulting terms results in \begin{align*} (a-2b)^2 + (b-2c)^2 + (c-2a)^2 &\ge 0 \\ a^2 - 4ab + 4b^2 + b^2 - 4bc + 4c^2 + c^2 - 4ac + 4a^2 &\ge 0 \\ 5a^2 + 5b^2 + 5c^2 &\ge 4ab + 4bc + 4ac \end{align*} Equality holds when $a-2b = b-2c = c-2a = 0.$ When that happens, $a = 2b = 4c = 8a,$ so $a=b=c=0.$

See Also

2003 Indonesia MO (Problems)
Preceded by
Problem 4
1 2 3 4 5 6 7 8 Followed by
Problem 6
All Indonesia MO Problems and Solutions
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