If $a>0$, $b>0$, $c>0$ and $$ a^2+b^2+c^2=1 $$ prove $$ \frac{a^3}{b^2+c}+\frac{b^3}{c^2+a}+\frac{c^3}{a^2+b} \geq \frac{\sqrt{3}}{\sqrt{3}+1} $$
$\frac{a^3}{b^2+c}+\frac{b^3}{c^2+a}+\frac{c^3}{a^2+b} \geq \frac{\sqrt{3}}{\sqrt{3}+1}$
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inequality
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0put $ in the title as well please. – 2017-02-17
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1i think you forgott some addtional conditions – 2017-02-17
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2Whenever you post any question on this site, do include your own research and efforts employed to tackle the problem. Also try to include the source of the question. I am advising you because otherwise your questions will be down-voted. – 2017-02-17
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0Try $a=-\frac12$, $b=\frac{\sqrt{3}}{2}$, and $c=0$. – 2017-02-17
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0it must be $$a>0,b>0,c>0$$ – 2017-02-17
1 Answers
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we have $$\frac{a^3}{b^2+c}+\frac{b^3}{c^2+a}+\frac{c^3}{a^2+b}\geq \frac{a^4}{ab^2+ac}+\frac{b^4}{bc^2+ab}+\frac{c^4}{a^2c+bc}\geq \frac{(a^2+b^2+c^2)^2}{ab^2+bc^2+ca^2+ab+bc+ca}=\frac{1}{ab^2+bc^2+ca^2+ac+bc+ca}$$ we have to Show that $$\sqrt{3}+1\geq \sqrt{3}(ab^2+bc^2+ca^2+ac+bc+ab)$$ we have $$1=a^2+b^2+c^2\geq ab+bc+ca$$ and $$(a^2c+b^2a+c^2b)^2\le (a^2+b^2+c^2)(a^2c^2+b^2c^2+a^2b^2)$$ by CS and $$3(a^2c^2+b^2c^2+c^2b^2)\le (a^2+b^2+c^2)^2$$ thus $$(a^2c+b^2a+c^2b)^2\le (a^2+b^2+c^2)^2(a^2c^2+b^2c^2+a^2b^2)\le \frac{1}{3}$$