https://artofproblemsolving.com/wiki/index.php?title=2007_Indonesia_MO_Problems/Problem_3&feed=atom&action=history2007 Indonesia MO Problems/Problem 3 - Revision history2024-03-28T14:02:24ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2007_Indonesia_MO_Problems/Problem_3&diff=118411&oldid=prevRockmanex3: Solution to Problem 3 (credit to vedran6) — niche inequality strat2020-02-26T03:50:46Z<p>Solution to Problem 3 (credit to vedran6) — niche inequality strat</p>
<p><b>New page</b></p><div>==Problem==<br />
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Let <math> a,b,c</math> be positive real numbers which satisfy <math> 5(a^2+b^2+c^2)<6(ab+bc+ca)</math>. Prove that these three inequalities hold: <math> a+b>c</math>, <math> b+c>a</math>, <math> c+a>b</math>.<br />
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==Solution (credit to vedran6)==<br />
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Assume that <math>a+b \le c</math>, so <math>c-a-b \ge 0</math>. Let <math>x = c-a-b</math>, making <math>x \ge 0</math>. Therefore, <math>c = a+b+x</math>.<br />
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By substitution, we have<br />
<cmath>\begin{align*}<br />
5(a^2 + b^2 + (a+b+x)^2) &< 6(ab + (a+b)(a+b+x)) \\<br />
5(a^2 + b^2 + a^2 + b^2 + x^2 + 2ab + 2ax + 2bx) &< 6(ab + a^2 + ab + ax + ab + b^2 + bx) \\<br />
5(2a^2 + 2b^2 + x + 2ab + 2ax + 2bx) &< 6(3ab + a^2 + ax + b^2 + bx) \\<br />
10a^2 + 10b^2 + 5x + 10ab + 10ax + 10bx &< 18ab+6a^2+6ax+6b^2+6bx \\<br />
4a^2 + 4b^2 + 5x - 8ab + 4ax + 4bx &< 0 \\<br />
4(a-b)^2 + x(5+4a+4b) &< 0<br />
\end{align*}</cmath><br />
Note that because <math>x, a, b \ge 0</math>, we must have <math>x(5+4a+4b) \ge 0</math>. Additionally, by the [[Trivial Inequality]], <math>4(a-b)^2 \ge 0</math>. Thus, we must have <math>4(a-b)^2 + x(5+4a+4b) \ge 0</math>, is a contradiction. Therefore, we must have <math>a+b > c</math>.<br />
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Because of symmetry in the equation, we can use similar steps to prove that <math> b+c>a</math> and that <math> c+a>b</math>.<br />
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==See Also==<br />
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{{Indonesia MO box|year=2007|num-b=2|num-a=4}}<br />
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[[Category:Intermediate Algebra Problems]]</div>Rockmanex3