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For positive integers Numbers $a, b, c $ prove that $$\frac{1+a^{2}}{1+b+c^{2}} +\frac{1+b^{2}}{1+c+a^{2}} +\frac{1+c^{2}}{1+a+b^{2}} \geq 2$$ This inequality above take long time to prove it and still couldn't complete it. How I can prove this inequality? Any hint will help. Thanks

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    $\begingroup$ I have to teach in a few minutes, but can you show it is increasing with respect to each variable, say by induction? Then $a=b=c=1$ would give the minimum. $\endgroup$ – Cheerful Parsnip Sep 16 '15 at 14:07
  • $\begingroup$ I would take the function calculate its gradient and calculate the critical points. Then i would use the Hessian to determine the type of critical point. Its is tedious but it think it is possible. $\endgroup$ – MrYouMath Sep 16 '15 at 14:25
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The above inequality holds for any positive real numbers $a,b,c$.

Indeed, denote $x=1+b+c^2,y=1+c+a^2,z=1+a+b^2$, the inequality becomes $$\frac{y-c}{x} + \frac{z-a}{y} + \frac{x-b}{z} \ge 2,$$ or equivalently $$\frac{y}{x} + \frac{z}{y} + \frac{x}{z} \ge 2 + \frac{c}{x} + \frac{a}{y} + \frac{b}{z}.$$ By AM-GM inequality: $$\frac{y}{x} + \frac{z}{y} + \frac{x}{z} \ge \sqrt[3]{\frac{y}{x} \frac{z}{y} \frac{x}{z}} =3.$$ It remains to prove $$\frac{c}{x} + \frac{a}{y} + \frac{b}{z} \le 1.$$ Since $x\ge 2c+b,y\ge 2a+c,z\ge 2b+a$ we only need to prove that $$\frac{c}{2c+b} + \frac{a}{2a+c} + \frac{b}{2b+a} \le 1,$$ which is equivalent to $$\left(\frac{c}{2c+b} - \frac{1}{2}\right) + \left(\frac{a}{2a+c} - \frac{1}{2}\right) + \left(\frac{b}{2b+a} - \frac{1}{2}\right) \le 1 - \frac{3}{2}$$ or $$\frac{b}{2c+b} + \frac{c}{2a+c} + \frac{a}{2b+a} \ge 1.$$ The last inequality is true by Cauchy-Schwarz inequality: $$\sum\frac{a}{2b+a} = \sum\frac{a^2}{2ab+a^2} \ge \frac{(a+b+c)^2}{\sum (2ab+a^2)}=1.$$ We are done.

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  • $\begingroup$ (+1 ) high solution sir. But could you please explain the last inequality. How we replace every $a, b, c $ in the numerator ? $\endgroup$ – user257567 Sep 16 '15 at 15:05
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    $\begingroup$ @user257567: Do you mean the Cauchy-Schwarz step? $\endgroup$ – Khue Sep 16 '15 at 15:09
  • $\begingroup$ @Khue I mean this sir.. $$\frac{b}{2c+b} + \frac{c}{2a+c} + \frac{a}{2b+a} \ge 1.$$ . How we get it? $\endgroup$ – user257567 Sep 16 '15 at 15:12
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    $\begingroup$ @user257567: $\sum \left( \frac{c}{2c+b} - \frac{1}{2} \right) \le -\frac{1}{2}$ equivalent to $\sum \frac{2c-(2c+b)}{2(2c+b)} \le -\frac{1}{2}$ equivalent to $\sum \frac{-b}{2(2c+b)} \le -\frac{1}{2}$ equivalent to $\sum \frac{b}{2c+b} \ge 1$. Is it clear enough? $\endgroup$ – Khue Sep 16 '15 at 15:17
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    $\begingroup$ @user257567: I'm glad I was able to help ;) $\endgroup$ – Khue Sep 16 '15 at 15:23
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Since $$\sum_{cyc}(1+a^2)(1+b+c^2)\sum_{cyc}\frac{1+a^2}{1+b+c^2} \ge \left(\sum_{cyc} (1+a^2)\right)^2,$$ we only need to show that $$\left(\sum_{cyc} (1+a^2)\right)^2\ge 2\sum_{cyc}(1+a^2)(1+b+c^2).\tag{1}$$ The LHS is $$\begin{aligned}2\sum_{cyc}(1+a^2+b+a^2b+c^2+a^2c^2)&= 6+4(a^2+b^2+c^2)+2(a^2b+b^2c+c^2a)\\ &+2(a+b+c)+2(a^2c^2+c^2b^2+b^2a^2). \end{aligned}\tag{2} $$ The RHS is $$(3+a^2+b^2+c^2)^2=9+6(a^2+b^2+c^2)+(a^2+b^2+c^2)^2.\tag{3}$$ Compare (2) and (3), we see that (1) is equivalent to $$a^4+b^4++c^4+2(a^2+b^2+c^2)+3\ge 2(a^2b+b^2c+c^2a)+2(a+b+c).\tag{4}$$ The last inequality is obvious from AM-GM: $$a^4+b^2\ge 2a^2b,\dots$$ and $$a^2+1\ge 2a,\dots$$ To finish the proof, we note that the equality happens in (4) only when $a=b=c=1$, which also yields equality $$\sum_{cyc}\frac{1+a^2}{1+b+c^2}=2.$$

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Let $1+a^2=x$, $1+b^2=y$ and $1+c^2=z$.

Hence, by AM-GM and C-S we obtain: $$\sum_{cyc}\frac{a^2+1}{1+b+c^2}\geq\sum_{cyc}\frac{a^2+1}{1+\frac{1+b^2}{2}+c^2}=\sum_{cyc}\frac{2x}{y+2z}=\sum_{cyc}\frac{2x^2}{xy+2xz}\geq$$ $$\geq\frac{2(x+y+z)^2}{\sum\limits_{cyc}(xy+2xz)}=\frac{2(x+y+z)^2}{3(xy+xz+yz)}=2+\frac{\sum\limits_{cyc}(x-y)^2}{3(xy+xz+yz)}\geq2.$$ Done!

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