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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

math110
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3 Answers3

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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

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

Macavity
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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
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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.

user1793963
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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.

Karuppiah Kannappan
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