First round – City competition

2.4. By using the condition $a+b+c=1$ and the inequality between arithmetic and geometric means, we have that $$\frac{a}{a+b^2}=\frac{a}{a(a+b+c)+b^2}=\frac{a}{a^2+b^2+ab+ac}\le \frac{a}{2ab+ab+ac}=\frac{1}{3b+c}.$$ By applying the inequality between harmonic and arithmetic means, it follows that $$\frac{4}{\frac{3}{b}+\frac{1}{c}}=\frac{4}{\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}}\le \frac{b+b+b+c}{4}=\frac{3b+c}{4},$$ i.e. $$\frac{a}{a+b^2}\le \frac{1}{3b+c}\le \frac{1}{16}\left(\frac{3}{b}+\frac{1}{c}\right).$$ Analogously, $$\frac{b}{b+c^2}\le \frac{1}{16}\left(\frac{3}{c}+\frac{1}{a}\right)\text{and}\frac{c}{c+a^2}\le \frac{1}{16}\left(\frac{3}{a}+\frac{1}{b}\right).$$ Finally, by adding the inequalities above, we have that $$\frac{a}{a+b^2}+\frac{b}{b+c^2}+\frac{c}{c+a^2}\le \frac{1}{16}\left(\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\right)=\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right).$$