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Difference between revisions of "2003 Indonesia MO Problems/Problem 5"

(Solution to Problem 5)
 
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and determine when the equality holds.
 
and determine when the equality holds.
  
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==Solution==
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==Solution 1==
  
 
By the [[Trivial Inequality]], <math>(a-2b)^2 \ge 0,</math> <math>(b-2c)^2 \ge 0,</math> and <math>(c-2a)^2 \ge 0.</math>  Adding all the equations, expanding the terms, and rearranging the resulting terms results in
 
By the [[Trivial Inequality]], <math>(a-2b)^2 \ge 0,</math> <math>(b-2c)^2 \ge 0,</math> and <math>(c-2a)^2 \ge 0.</math>  Adding all the equations, expanding the terms, and rearranging the resulting terms results in
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5a^2 + 5b^2 + 5c^2 &\ge 4ab + 4bc + 4ac
 
5a^2 + 5b^2 + 5c^2 &\ge 4ab + 4bc + 4ac
 
\end{align*}</cmath>
 
\end{align*}</cmath>
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Equality holds when <math>a-2b = b-2c = c-2a = 0.</math>  When that happens, <math>a = 2b = 4c = 8a,</math> so <math>a=b=c=0.</math>
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Equality holds when <math>a-2b = b-2c = c-2a = 0</math>  When that happens, <math>a = 2b = 4c = 8a,</math> so <math>a=b=c=0.</math>
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==Solution 2==
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<math> 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</math>
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Equality holds when <math>a^2=b^2=c^2=0, a=b=c=0</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 03:47, 8 November 2022

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 $a^2=b^2=c^2=0, a=b=c=0$.

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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