How does one prove $$\frac {a^{\frac{3}{2}}}{a+b}+\frac {b^{\frac{3}{2}}}{b+c}+\frac {c^{\frac{3}{2}}}{c+a} \ge \frac {a+b+c}{\sqrt 2}$$ where $a, b, c$ are positive?
I tried different formats. Where could that $\sqrt2$ come from?
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Please use Mathjax https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Albus Dumbledore Oct 30 '20 at 03:00
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Does your square root apply to the whole fraction? i.e. $\dfrac {\sqrt {a^3}} {a+b}$ or $\sqrt{\dfrac {a^3}{a+b}}$? – player3236 Oct 30 '20 at 03:04
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Sorry, I'll learn to use Mathjax.. No, sqrt only applies to a^3, but not (a+b). Thanks. – renmom Oct 30 '20 at 03:06
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If the square root applies to the whole fraction, there is a variety of solutions in this post: https://artofproblemsolving.com/community/c6h226502p1256980. However if it does not, take $a=b=c=1$, then $3/2 < 3/\sqrt2$ contradicts your statement. – player3236 Oct 30 '20 at 03:11
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The question seems wrong due to scaling. Do you have any constraints on $a,b,c$ other than positivity? – Oct 30 '20 at 03:12
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1You all are good at this! I believe the original problem print was wrong. The sqrt applies to all. The aops solution looks good. Thanks so much. – renmom Oct 30 '20 at 03:14