Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=1$. Prove that $$ \sum_{cyc}\frac{b\sqrt{c}}{a(\sqrt{3c}+\sqrt{ab})}\geq \frac{3\sqrt{3}}{4} .$$

My attempt :

Substitute $a=x^2, b=y^2, c=z^2$, where $x^2+y^2+z^2=1$

$\displaystyle\sum_{cyc}\frac{b\sqrt{c}}{a(\sqrt{3c}+\sqrt{ab})}= \sum_{cyc}\frac{y^2z}{\sqrt{3}x^2z+x^3y} $

Please suggest on how to proceed using basic inequalities if possible.

  • $\begingroup$ what if try direction cosines substitution? $\endgroup$ – serg_1 Jul 30 '17 at 9:11
  • $\begingroup$ @serg_1, I didn't use to solve inequality using Trigonometry, I know just its basic. $\endgroup$ – carat Jul 31 '17 at 8:43

We need to prove that $$\sum_{cyc}\frac{b^2c}{a^2(\sqrt{3(a^2+b^2+c^2)}c+ab)}\geq\frac{3}{4}\sqrt{\frac{3}{a^2+b^2+c^2}}$$ for positives $a$, $b$ and $c$. $$\sum_{cyc}\frac{b^2c}{a^2(\sqrt{3(a^2+b^2+c^2)}c+ab)}\geq\frac{3}{4}\sqrt{\frac{3}{a^2+b^2+c^2}}$$ or $$\sum_{cyc}\frac{6b^2c}{a^2(2\sqrt{3(a^2+b^2+c^2)}3c+6ab)}\geq\frac{3}{4}\sqrt{\frac{3}{a^2+b^2+c^2}}.$$ Now, by AM-GM and C-S we obtain $$\sum_{cyc}\frac{6b^2c}{a^2(2\sqrt{3(a^2+b^2+c^2)}3c+6ab)}\geq\sum_{cyc}\frac{6b^2c}{a^2(3(a^2+b^2+c^2)+9c^2+6ab)}=$$ $$=2\sum_{cyc}\frac{b^2c}{a^2(4c^2+a^2+b^2+2ab)}=2\sum_{cyc}\frac{b^4c^4}{a^2b^2c^2(4c^3+a^2c+b^2c+2abc)}\geq$$ $$\geq\frac{2(a^2b^2+a^2c^2+b^2c^2)^2}{a^2b^2c^2\sum\limits_{cyc}(4a^3+a^2b+a^2c+2abc)}.$$ Id est, it's enough to prove that $$8(a^2b^2+a^2c^2+b^2c^2)^2\sqrt{\frac{a^2+b^2+c^2}{3}}\geq3 a^2b^2c^2\sum\limits_{cyc}(4a^3+a^2b+a^2c+2abc).$$ Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Hence, we need to prove that $$9(9v^4-6uw^3)^2\sqrt{3u^2-2v^2}\geq3w^6(4(27u^3-27uv^2+3w^3)+9uv^2-3w^3+6w^3$$ or $f(w^3)\geq0$, where $$f(w^3)=3(3v^4-2uw^3)^2\sqrt{3u^2-2v^2}-(36u^3-33uv^2+5w^3)w^6,$$ which is decreasing function, which says that it's enough to prove the last inequality

for a maximal value of $w^3$, which happens for equality case of two variables.

Since the last inequality is homogeneous, we can assume $b=c=1$, which gives $$8(2a^2+1)^2\sqrt{\frac{a^2+2}{3}}\geq3a^2(4(a^3+2)+2(a^2+a+1)+6a)$$ or $g(a)\geq0$, where $$g(a)=2\ln(2a^2+1)+\frac{1}{2}\ln(a^2+2)-2\ln{a}-\ln(2a^3+a^2+4a+5)+2\ln2-\frac{3}{2}\ln3.$$ But $$g'(a)=\frac{(a-1)(2a^5+2a^4+29a^3+17a^2+44a+20)}{a(a^2+2)(2a^2+1)(2a^3+a^2+4a+5)},$$ which gives $a_{min}=1$ and since $g(1)=0,$ we are done!

  • $\begingroup$ This problem is hard. Thank you for your help. $\endgroup$ – carat Jul 31 '17 at 8:44
  • $\begingroup$ @carat You are welcome! $\endgroup$ – Michael Rozenberg Jul 31 '17 at 9:34

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