Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Show that $$\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\geq \frac{3}{2}.$$
Problem
Source: Moldova TST 2018, b6
Tags: inequalities, three variable inequality, algebra
06.04.2018 17:48
Snakes wrote: Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Prove: $$\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\geq \frac{3}{2}$$ Notice that $\frac{a}{1+b^2}=a-\frac{ab^2}{1+b^2}$. Thus, it's suffice to prove that $$\frac{3}{2}\geq \sum_{cyc} \frac{ab^2}{1+b^2}$$ But by AM-GM, $\frac{ab^2}{1+b^2}\leq \frac{ab}{2}$. Also, $ab+bc+ca\leq \frac{(a+b+c)^2}{3}=3$. So, we are done!
06.04.2018 18:21
Snakes wrote: Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Prove: $$\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\geq \frac{3}{2}$$ See also here https://artofproblemsolving.com/community/c6h1199404p5898263
06.04.2018 18:21
indeed very easy for TST
28.04.2018 07:38
$P(a,b,c)$ be the lhs of the inequality. $a+b+c=3$ we can establish equality obvious $a=b=c=1$ and take an example $(a, b, c) =(2,\frac{1}{2},\frac{1}{2})$ then obvious $P(a,b,c)\ge \frac{3}{2}$ holds also $a, b, c$ are positive. $P(a,b,c)\ge 9\sum_{\text{cyc}}(\frac{\frac{1+b^2}{a}}{a})^{–1}$ We have used the $AM\ge HM$ inequality . But $\frac{a^2}{x}+\frac{b^2}{y}\ge \frac{(a+b)^2}{x+y}$ holds for $a, b$ are in $R$ $x, y>0$.therefore This rhs $=Q(a,b,c)$ (say) Now, we know that $\frac{1^2}{a}+\frac{1^2}{b}+\frac{1^2}{c}\ge \frac{(1+1+1)^2}{a+b+c}=3$ Also we have the inequality $\sum_{\text{cyc}}\frac{b^2}{a}\ge \frac{(a+b+c)^2}{a+b+c}=(a+b+c)=3$ Therefore we can say comfortably $P(a,b,c)\ge 9\frac{1}{3+3}=\frac{3}{2}$ Thus $P(a,b,c)\ge \frac{3}{2}$ This inequality can be solved by Titus lemma also.
15.06.2021 03:19
Snakes wrote: Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Show that $$\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\geq \frac{3}{2}.$$ Let $a,b,c>0$ and $a+b+c=3.$ Prove that$$\frac{a}{k+b^2}+\frac{b}{k+c^2}+\frac{c}{k+a^2}\ge\frac{3}{k+1}$$Where $0\le k\le 3+2\sqrt{3}.$ (Victoria_Discalceata1) h For all positive real number $a,b,c$ such that $a+b+c=3$. Prove that: $$\frac{a}{1+b^3}+\frac{b}{1+c^3}+\frac{c}{1+a^3} \geq \frac{3}{2}$$
15.06.2021 03:31
Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Prove or disprove $$\frac{a}{1+b^n}+\frac{b}{1+c^n}+\frac{c}{1+a^n}\geq \frac{3}{2}.$$Where $ n\in N^+.$