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For $a,b,c>0$ and $ab+bc+ca=3$. Prove that $$P=\frac{a^2}{\sqrt{3b^2+bc}}+\frac{b^2}{\sqrt{3c^2+ca}}+\frac{c^2}{\sqrt{3a^2+ab}}\ge \frac{3}{2}$$


By AM-GM $\frac{a^3}{2\sqrt{3ab^2+abc}}+\frac{a^3}{2\sqrt{3ab^2+abc}}+\frac{3ab^2+abc}{16}\ge 3\sqrt[3]{\frac{a^6}{64}}=\frac{3a^2}{4}$

$\Rightarrow P+\sum _{cyc}\frac{3ab^2+abc}{16}\ge \frac{3\left(a^2+b^2+c^2\right)}{4}\ge \frac{9}{4}$

Need prove $\sum _{cyc}\frac{3ab^2+abc}{16}\ge \frac{9}{4}-\frac{3}{2}=\frac{3}{4}$

Help

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By Holder $$\left(\sum_{cyc}\frac{a^2} {3b^2+bc}\right)^2\sum_{cyc}a^2(3b^2+bc)\geq(a^2+b^2+c^2)^3.$$ Thus, it remains to prove that $$4(a^2+b^2+c^2)^3\geq3(ab+ac+bc)\sum_{cyc}a^2(3b^2+bc),$$ which is true because $$a^2+b^2+c^2\geq ab+ac+bc$$ and $$4(a^2+b^2+c^2)^2\geq12\sum_{cyc}a^2b^2=3\sum_{cyc}4a^2b^2\geq3\sum_{cyc}(3a^2b^2+a^2bc).$$ Done!

Another way. $$a+b+c=\sqrt{(a+b+c)^2}=\sqrt{\sum_{cyc}(a^2+2bc)}\geq\sqrt{\sum_{cyc}(ab+2bc)}=3.$$ Thus, by AM-GM and Holder we obtain: $$\sum_{cyc}\frac{a^2}{\sqrt{3b^2+bc}}=\sum_{cyc}\frac{4a^2}{2\sqrt{4b(3b+c)}}\geq\sum_{cyc}\frac{4a^2}{4b+3b+c}=$$ $$=\sum_{cyc}\frac{4a^3}{7ab+ac}\geq\frac{4(a+b+c)^3}{3\sum\limits_{cyc}(7ab+ac)}=\frac{(a+b+c)^3}{18}\geq\frac{27}{18}=\frac{3}{2}.$$

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  • $\begingroup$ first ano :why $4(a^2+b^2+c^2)^2\geq12\sum_{cyc}a^2b^2$ $\endgroup$ – Word Shallow Jun 30 '17 at 13:31
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    $\begingroup$ @Word Shallow It's $(a^2+b^2+c^2)^2\geq3(a^2b^2+a^2c^2+b^2c^2)$ or $\sum\limits_{cyc}(a^2-b^2)^2\geq0$. $\endgroup$ – Michael Rozenberg Jun 30 '17 at 13:33

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