parmenides51 wrote: Let $a, b, c$ be positive reals. Prove that $$\left(\frac{a}{a+b}\right)^2+\left(\frac{b}{b+c}\right)^2+\left(\frac{c}{c+a}\right)^2\ge \frac34$$ https://artofproblemsolving.com/community/c6h81368p466211 https://artofproblemsolving.com/community/c6h303978p1644390 Let $a, b, c$ be reals such that $(a+b)(b+c)(c+a)\neq 0. $ Prove that $$\left(\frac{a}{a+b}\right)^2+\left(\frac{b}{b+c}\right)^2+\left(\frac{c}{c+a}\right)^2+\frac{4abc}{(a+b)(b+c)(c+a)}\geq 1$$(Crux (9)2021 Problem 4670)

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parmenides51 wrote: Let $a, b, c$ be positive reals. Prove that $$\left(\frac{a}{a+b}\right)^2+\left(\frac{b}{b+c}\right)^2+\left(\frac{c}{c+a}\right)^2\ge \frac34$$ $$\sum_{cyc}\left(\frac{a}{a+b}\right)^2=\left(\frac{1}{1+\frac{b}{a}}\right)^2+\left(\frac{1}{1+\frac{c}{b}}\right)^2+\left(\frac{c}{c+a}\right)^2\geq$$$$\geq\frac{1}{1+\frac{b}{a}\cdot\frac{c}{b}}+\left(\frac{c}{c+a}\right)^2=\frac{3}{4}+\frac{(a-c)^2}{4(a+c)^2}\geq\frac{3}{4}.$$

Use the identity: $$ \frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}=\frac{(a-b)(b-c)(a-c)}{(a+b)(b+c)(a+c)} $$Multiply the two sides by $4$: $$\sum{(\frac{2a}{a+b})^2}=\sum{(\frac{a-b}{a+b} +1)^2}=\sum{(\frac{a-b}{a+b})^2}+2\sum{\frac{a-b}{a+b}} +3 \geq 3$$$$\leftrightarrow\sum{(\frac{a-b}{a+b})^2}+2\sum{\frac{a-b}{a+b}} \geq 0$$Set $x=\sum{\frac{a-b}{a+b}}=-3+2\sum{\frac{a}{a+b}}>-3$ . Then we have: $$\sum{(\frac{a-b}{a+b})^2}+2\sum{\frac{a-b}{a+b}} \geq 3(\prod{\frac{a-b}{a+b}})^{\frac{2}{3}} + 2x =3x^{\frac{2}{3}} +2x$$Its easy to see for $x>-3$, $3x^{\frac{2}{3}} +2x \geq 0$.