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If $a$, $b$ and $c$ are positive then $\sum\limits_{cyc} \frac{a^{2}}{b^{2}+c^{2}+bc}\geq 1$.

I tried to solve this problem by C-S. But I can't sovle it. Things I have done so far: $\sum\limits_{cyc} \frac{a^{2}}{b^{2}+c^{2}+bc}\geq \frac{(\sum\limits_{cyc}a)^2}{2.\sum\limits_{cyc}a^2+\sum\limits_{cyc}ab}\geq \frac{(\sum\limits_{cyc}a)^2}{3.\sum\limits_{cyc}a^2}=\frac{\sum\limits_{cyc}a^2+2.\sum\limits_{cyc}ab}{3.\sum\limits_{cyc}a^2}=\frac{1}{3}+\frac{2\sum\limits_{cyc}ab}{3.\sum\limits_{cyc}a^2}\geq ?$

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    $\begingroup$ What is the summation over? What exactly does $cyc$ mean. $\endgroup$ – Sean Haight Aug 7 '18 at 5:02
  • $\begingroup$ Have you tried exploiting homogeneity? e.g. assume $a+b+c=1$. $\endgroup$ – Alex R. Aug 7 '18 at 5:08
  • $\begingroup$ Alex R, I haven't. Can you help me? Sorry, I am not good at English so it might be difficult for me to explain the things that I didn't understand. $\endgroup$ – Shizumi Aoki Aug 7 '18 at 5:12
  • $\begingroup$ @SeanHaight possibly it means that the terms of the sum can be expressed by cycling $a$, $b$ and $c$. That is, $a \mapsto b$, $b \mapsto c$, $c \mapsto a$ gives an expression which is another term of the sum. Repeating this process gives all the terms. $\endgroup$ – Brahadeesh Aug 9 '18 at 7:37
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By C-S and Muirhead $$\sum_{cyc}\frac{a^2}{b^2+c^2+bc}=\sum_{cyc}\frac{a^4}{a^2b^2+a^2c^2+a^2bc}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^2b^2+a^2c^2+a^2bc)}=$$ $$=\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(2a^2b^2+a^2bc)}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(2a^2b^2+a^4)}=1.$$ Also, by Holder and AM-GM: $$\sum_{cyc}\frac{a^2}{b^2+c^2+bc}=\sum_{cyc}\frac{a^3}{b^2a+c^2a+abc}\geq\frac{(a+b+c)^3}{3\sum\limits_{cyc}(ab^2+ac^2+abc)}=$$ $$=\frac{(a+b+c)^3}{\sum\limits_{cyc}(abc+3a^2b+3a^2c+2abc)}\geq\frac{(a+b+c)^3}{\sum\limits_{cyc}(a^3+3a^2b+3a^2c+2abc)}=1.$$

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Very strightforward solution:

Given inequality is equivalent to:

$$\frac{a^{2}}{b^{2}+c^{2}+bc}+\frac{b^{2}}{c^{2}+a^{2}+ca}+\frac{c^{2}}{a^{2}+b^{2}+ab}-1\geq 0$$

or:

$$a^6+a^5 b+a^5 c-a^3 b^3-a^3 b^2 c-a^3 b c^2-a^3 c^3-a^2 b^3 c-a^2 b c^3+a b^5-a b^3 c^2-a b^2 c^3+a c^5+b^6+b^5 c-b^3 c^3+b c^5+c^6 \ge 0$$

If you introduce $T[p,q,r]=\sum_{sym}a^pb^qc^r$ you get:

$$\frac12T[6,0,0]+T[5,1,0]-T[3,2,1]-\frac12 T[3,3,0]\ge0$$

...which is obviously true because:

$$T[6,0,0]\ge T[3,3,0],\quad T[5,1,0]\ge T[3,2,1]$$

(Muirhead)

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