Iran TST 2012,Exam2,Day2,P4

For discussing Olympiad Level Algebra (and Inequality) problems
tanmoy
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Joined:Fri Oct 18, 2013 11:56 pm
Location:Rangpur,Bangladesh
Iran TST 2012,Exam2,Day2,P4

Unread post by tanmoy » Sat Oct 15, 2016 7:44 pm

For positive reals $a,b$ and $c$ with $ab+bc+ca=1$, show that
\[\sqrt{3}({\sqrt{a}+\sqrt{b}+\sqrt{c})\le \frac{a\sqrt{a}}{bc}+\frac{b\sqrt{b}}{ca}+\frac{c\sqrt{c}}{ab}.}\]
"Questions we can't answer are far better than answers we can't question"

Mehedi Hasan Nowshad
Posts:12
Joined:Sat Jun 13, 2015 1:46 pm
Location:Halishahar, Chittagong

Re: Iran TST 2012,Exam2,Day2,P4

Unread post by Mehedi Hasan Nowshad » Sun Nov 06, 2016 10:37 pm

My solution
Let $a \ge b \ge c > 0$. Then by applying chebyshev, we get,
\[ \sum_{cyc}^{} \dfrac{a\sqrt{a}}{bc} \ge \dfrac{1}{3} \left( \sum_{cyc}^{} \dfrac{a}{bc} \right)\left( \sum_{cyc}^{} \sqrt{a} \right) \]

Then it's enough to prove that
\begin{align*}
\dfrac{1}{3} \left( \sum_{cyc} \dfrac{a}{bc} \right) &\ge \sqrt{3} \\
or, \sum a^2 &\ge 3\sqrt{3} abc \\
or, \left( \sum a^2 \right) \left(\sqrt{\sum ab} \right) &\ge 3\sqrt{3} abc \\
\end{align*}

Which is clearly true by AM-GM and we are done :D
"Failure is simply the opportunity to begin again, this time more intelligently."
- Henry Ford

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