2003 Indonesia MO Problems/Problem 5

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 1

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 $a=b=c=0.$

Solution 2

$5a^2 + 5b^2 + 5c^2 \ge 4a^2 + 4b^2 + 4c^2 = 4ab + 4bc + 4ac+ 2(a-b)^2 + 2(b-c)^2 + 2(c-a)^2 \ge 4ab + 4bc + 4ac$

Equality holds when $a^2=b^2=c^2=0, a=b=c=0$ .

2003 Indonesia MO (Problems)
Preceded by Problem 4	1 • 2 • 3 • 4 • 5 • 6 • 7 • 8	Followed by Problem 6
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2003 Indonesia MO Problems/Problem 5

Contents

Problem

Solution 1

Solution 2

See Also