parmenides51 wrote: Let $a,b,c$ be positive real numbers where $ab+bc+ca = 3$. Prove that $$\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\geq\dfrac{3} {2}.$$(dektep) http://artofproblemsolving.com/community/c6h1087398p4816912

parmenides51 wrote: Let $a,b,c$ be positive real numbers where $ab+bc+ca = 3$. Prove that $$\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\geq\dfrac{3} {2}.$$(dektep) pqr method. Let $p=a+b+c, q = ab + bc + ca = 3, r = abc$. Full expanding yields $p^2 + 2pr - 3r^2-12 \ge 0$ which is true (easy to prove using $q^2 \ge 3pr$ and $p^2 \ge 3q$).

parmenides51 wrote: Let $a,b,c$ be positive real numbers where $ab+bc+ca = 3$. Prove that $$\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\geq\dfrac{3} {2}.$$(dektep) non pqr method WLOG, assume $a\ge b\ge c$. Then $ab\ge 1$ and $1\ge c$ We prove the following inequality: $$\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}\geq\dfrac{2} {ab+1}.$$This is equivalent to $(ab-1)(a-b)^2\ge0$ Apply this, we need to prove $\dfrac{2} {ab+1}+\dfrac{1} {c^2+1}\ge\dfrac{3}{2}$ Notice that $ab+1=4-c(a+b)$ And $a+b+c\ge3$ -> $a+b\ge3-c$ => $4-c(3-c)\ge4-c(a+b)$ We need to prove: $\dfrac {2}{c^2-3c+4}+\dfrac{1}{c^2+1}\ge\dfrac{3}{2}$ Or $\dfrac {3c(1-c)}{2(c^2+1)(c^2-3c+4}\ge0$ (true because c<=1) Equality reached at $a=b=c=1$

Nice. Actually, yesterday I also wrote down another proof which is similar to yours. WLOG, assume that $ab \ge 1$. We have $$\frac{1}{1 + a^2} + \frac{1}{1 + b^2} - \frac{2}{ab + 1} = \frac{(ab-1)(a-b)^2}{(a^2+1)(b^2+1)(ab+1)}\ge 0.$$ From $ab + bc + ca = 3$, we have $ab + 2\sqrt{ab} \cdot c \le 3$ which results in $\sqrt{ab} \le \sqrt{c^2+3} - c$. It suffices to prove that $$\frac{2}{(\sqrt{c^2+3} - c)^2 + 1} + \frac{1}{c^2+1} \ge \frac32$$which is true. We are done.

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That little comparison at the end using AM-GM is much stronger than mine . How did you get to know this inequality btw$$\frac{1}{1 + a^2} + \frac{1}{1 + b^2} \ge \frac{2}{ab + 1}$$

Schur-Schwartz wrote:

That little comparison at the end using AM-GM is much stronger than mine . How did you get to know this inequality btw$$\frac{1}{1 + a^2} + \frac{1}{1 + b^2} \ge \frac{2}{ab + 1}$$ I think it is a known trick. There are some tricks to deal with similar things. For example, many years ago, I knew this: for $x, y \ge 0; \ xy\le 4$, $$\frac{1}{\sqrt{1+x}} + \frac{1}{\sqrt{1+y}} \le \frac{2}{\sqrt{1+\sqrt{xy}}}.$$ Also, under some condition, we have $$\frac{1}{y^2 + 1} + \frac{1}{z^2 + 1} \ge 1 + \frac{1}{1 + (y + z)^2}.$$