# Prove $\frac{a}{b^{2}+1} + \frac{b}{c^{2}+1} + \frac{c}{a^{2} + 1} \ge \frac{3}{2}$

$$a,b,c > 0$$ and $$a+b+c=3$$, prove

$$\frac{a}{b^{2} + 1} + \frac{b}{c^{2}+1} + \frac{c}{a^{2}+1} \ge 3/2$$

Attempt:

Notice that by AM-Gm

$$\frac{a}{b^{2} + 1} + \frac{b}{c^{2}+1} + \frac{c}{a^{2}+1} \ge 3\frac{\sqrt[3]{abc}}{\sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)}}$$

Now, AM-GM again

$$a^{2}+b^{2}+c^{2} + 3 \ge 3 \sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)} ... (1)$$

Then $$a+b+c = 3 \ge 3 \sqrt[3]{abc} \implies 1 \ge \sqrt[3]{abc}$$. Also

$$a^{2} + b^{2} + c^{2} \ge 3 \sqrt[3]{(abc)^{2}}$$ multiply by $$1 \ge \sqrt[3]{abc}$$ and will get

$$a^{2} + b^{2} + c^{2} \ge 3 abc ... (2)$$

subtract $$(1)$$ with $$(2)$$ and get

$$3 \ge 3 \sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)} - 3 abc$$ $$3 + 3 abc \ge \sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)}$$ $$\frac{3abc}{\sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)}} \ge 1 - \frac{3}{\sqrt[3]{(a^{2}+1)(b^{2}+1)(c^{2}+1)}}$$

How to continue..?

• more often than not, the minimum or maximum comes at the value where they are equal to each other, and here when a=b=c=1, though its not related to your question directly – George carlin Sep 18 '19 at 9:40
• Possibly related : math.stackexchange.com/questions/2606354/… – Arnaud D. Sep 18 '19 at 13:13

$$\sum_{cyc}\frac{a}{b^2+1}=3+\sum_{cyc}\left(\frac{a}{b^2+1}-a\right)=3-\sum_{cyc}\frac{ab^2}{b^2+1}\geq$$ $$\geq3-\sum_{cyc}\frac{ab^2}{2b}=3-\frac{1}{2}(ab+ac+bc).$$ Can you end it now?
Since by your work $$3-\frac{1}{2}(ab+ac+bc)=3-\frac{1}{2}\cdot\frac{9-a^2-b^2-c^2}{2},$$ it's enough to prove that $$3-\frac{1}{2}\cdot\frac{9-a^2-b^2-c^2}{2}\geq\frac{3}{2}$$ or $$a^2+b^2+c^2\geq3,$$ which is true by C-S: $$a^2+b^2+c^2=\frac{1}{3}(1^2+1^2+1^2)(a^2+b^2+c^2)\geq\frac{1}{3}(a+b+c)^2=3.$$
• I change to $(ab+ac+bc) = \frac{9 - (a^{2} + b^{2} + c^{2})}{2}$, but then how to prove $a^{2}+b^{2}+c^{3} \ge 3$? – Arief Sep 19 '19 at 3:07
Since for $$x>0$$ we have (just draw a graph for $${1\over 1+x^2}$$ and a tangent at $$x=1$$) $${1\over 1+x^2}\geq -{1\over 2}x+1$$ it is enough to check if $$-{1\over 2}(ab+bc+ca)+3\geq {3\over 2}$$ i.e. $$3\geq ab+bc+ca$$ is true?
Since $$a^2+b^2+c^2\geq ab+bc+ca$$ that is easy to verfy. :)