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Let $a,b,c$ be nonnegative real numbers such that $a+b+c=3$, Prove that

$$ \sqrt{\frac{a}{a+3b+5bc}}+\sqrt{\frac{b}{b+3c+5ca}}+\sqrt{\frac{c}{c+3a+5ab}}\geq 1.$$

This problem is from http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=555716

@Calvin Lin Thank you

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It might help to substitute $a=3x^2$, $b=3y^2$ and $c=3z^2$. Equality holds for $a=b=c=1$. –  Axel Kemper Sep 26 '13 at 8:40

3 Answers 3

up vote 3 down vote accepted

Let $\displaystyle A = \sqrt{\frac{a}{a+3b+5bc}}+\sqrt{\frac{b}{b+3c+5ca}}+\sqrt{\frac{c}{c+3a+5ab}}$ and $\displaystyle B = \sum_{cyc}a^2(a+3b+5bc)$.

Then by Hölder's inequality we have $A^2B \ge (a+b+c)^3 = 27$.

So it is sufficient to prove that $B \le 27$

$$B = \sum_{cyc}a^3 + 3 \sum_{cyc}a^2b+5\sum_{cyc}a^2bc$$

As $\displaystyle \sum_{cyc}ab^2 \ge 3abc$ by AM-GM, we have

$$B \le \left(\sum_{cyc}a^3 + 3 \sum_{cyc}a^2b + 3 \sum_{cyc}ab^2 + 6abc\right) - 15abc + 5\sum_{cyc}a^2bc \\ = (a+b+c)^3 - 5abc (3-\sum_{cyc}a) = 27$$

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+1; Nice answer; but for Holder inequality I got $(\sum a^{2/3})3$. Just want to be sure that everything is fine. –  Arash Sep 26 '13 at 19:28
    
@Arash You are absolutely right, my bad. Have corrected it now. –  Macavity Sep 26 '13 at 19:45
    
By the way, $5\sum_{cyc}a^2bc=15abc$ so the last part you have equality. Also the last term is $5\sum_{cyc}abc(a - 1)$, isn't it? –  Arash Sep 26 '13 at 20:01
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@Arash All that is accounted for in the last edit. Thanks anyway. –  Macavity Sep 26 '13 at 20:11
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You are welcome! I hope we can turn this website into something with more hospitable comments rather than hostile comments from arrogant people. :-) –  Arash Sep 26 '13 at 20:13

We know the one result if sum of two numbers $a$ and $b$ is constant, the max. of $ab$ occurs when both $a$ and $b$ equal. Eg. if $a+b=6$, then $\max\{ab\}=9$. Use this result, we get proof.

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You don't know inequality... –  user94270 Sep 26 '13 at 9:07
    
I know well. I give hint only. using this hint, you prove this result. –  Karuppiah Kannappan Sep 26 '13 at 11:42

Since $a+b+c=3$ you can reduce $f(a,b,c)$ to $f(3-b-c, b, c)$. With only 2 variables you can make a 3D plot and see if it gets below $1$ for $b+c\le3$. Not an elegant proof, but it works. If nothing else it will at least show you where the edge cases are.

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Are you implying that looking at a graph is a real proof ? –  user37238 Sep 26 '13 at 9:51
    
Sorry, I worded that badly, I meant it is not a proof and not very elegant, but it can be helpfull to see how the function behaves. –  user1793963 Sep 26 '13 at 11:50

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