Cyclic Olympiad Inequality

Question

Given $a^2+b^2+c^2=1$ Prove $\sum_\text{cyc} \frac{1}{6ab+c^2}-\frac{1}{2+c^2}$ is nonnegative I have tried substituting 1 with $a^2+b^2+c^2$, but nothing is working. I’m trying to reduce it into a one-variable inequality. Then use Jensen or derivatives or n-1 EV

Answer 1

For positive variables SOS helps.

We need to prove that $$\sum_{cyc}\left(\frac{1}{c^2+6ab}-\frac{1}{2a^2+2b^2+3c^2}\right)\geq0$$ or $$\sum_{cyc}\frac{2a^2+2b^2+2c^2-6ab}{(c^2+6ab)(c^2+2)}\geq0$$ or $$\sum_{cyc}\frac{(c-a)(3b+c-2a)-(b-c)(3a+c-2b)}{(c^2+6ab)(c^2+2)}\geq0$$ or $$\sum_{cyc}(a-b)\left(\frac{3c+a-2b}{(a^2+6bc)(a^2+2)}-\frac{3c+b-2a}{(b^2+6ac)(b^2+2)}\right)\geq0$$ or $$\sum_{cyc}(a-b)^2(2a^4+a^3b+a^2b^2+2b^4-3a^3c+9a^2bc+9b^2ac-3b^3c+4a^2+2ab+4b^2+6ac+6bc+36c^2)(c^2+6ab)(c^2+2)\geq0,$$ which is true because $$2a^4+a^3b+a^2b^2+2b^4-3a^3c+9a^2bc+9b^2ac-3b^3c+4a^2+2ab+4b^2+6ac+6bc+36c^2=$$ $$=2a^4+a^3b+a^2b^2+2b^4-3a^3c+9a^2bc+9b^2ac-3b^3c+(4a^2+2ab+4b^2+6ac+6bc+36c^2)(a^2+b^2+c^2)>0.$$

Answer 2

Also, we can use the following way.

By AM-GM $$\sum_{cyc}\frac{1}{c^2+6ab}\geq\sum_{cyc}\frac{1}{c^2+3a^2+3b^2}.$$ Thus, we need to prove that $$\sum_{cyc}\frac{1}{c+3a+3b}\geq\sum_{cyc}\frac{1}{3c+2a+2b}$$ for positives $a$, $b$ and $c$, which follows from the integration of the following obvious inequality: $$\sum_{cyc}x^3y^3z\geq\sum_{cyc}x^3y^2z^2,$$ where $x$, $y$ and $z$ are positives.

Indeed, let $x=t^a$, $y=t^b$ and $z=t^c$, where $t>0$. Thus, $$\sum_{cyc}x^3y^3z\geq\sum_{cyc}x^3y^2z^2$$ gives $$\sum_{cyc}t^{3a+3b+c}\geq\sum_{cyc}t^{3a+2b+2c}$$ or $$\sum_{cyc}t^{3a+3b+c-1}\geq\sum_{cyc}t^{3a+2b+2c-1},$$ which gives $$\int\limits_{0}^1\sum_{cyc}t^{3a+3b+c-1}dx\geq\int_0^1\sum_{cyc}t^{3a+2b+2c-1}dx$$ or $$\sum_{cyc}\int\limits_{0}^1t^{3a+3b+c-1}dx\geq\sum_{cyc}\int_0^1t^{3a+2b+2c-1}dx$$ or $$\sum_{cyc}\frac{1}{3a+3b+c}\geq\sum_{cyc}\frac{1}{3a+2b+2c}$$ and we are done!

Answer 3

Also, we can use C-S.

Indeed, after using AM-GM it's enough to prove that $$\sum_{cyc}\frac{1}{c+3a+3b}\geq\sum_{cyc}\frac{1}{3c+2a+2b}$$ for positives $a$, $b$ and $c$ and by C-S we obtain: $$\sum_{cyc}\frac{1}{3c+2a+2b}=2\sum_{cyc}\frac{1}{6c+4a+4b}=2\sum_{cyc}\frac{1}{3c+3a+b+3c+3b+a}\leq$$ $$\leq\frac{2}{(1+1)^2}\sum_{cyc}\left(\frac{1^2}{3c+3a+b}+\frac{1^2}{3c+3b+a}\right)=\sum_{cyc}\frac{1}{c+3a+3b}.$$