We need to equivalently prove $$\sum\frac{(a+b)^2}{4+(a+b)^2}\geq\frac32$$By C-S, $$LHS\geq\frac{4(a+b+c)^2}{12+2(a^2+b^2+c^2+ab+bc+ca)}$$So it suffices to prove $8(a+b+c)^2\geq 36+6(a^2+b^2+c^2+ab+bc+ca)$ which is true by $a^2+b^2+c^2\geq ab+bc+ca$

nice proof .

parmenides51 wrote: Positive real numbers $a, b,c$ satisfy $ab + bc+ ca = 3$. Prove the inequality $$\frac{1}{4+(a+b)^2}+\frac{1}{4+(b+c)^2}+\frac{1}{4+(c+a)^2}\le \frac{3}{8}$$ Azerbaijan BMO 2016 Preparation Exam Let $ a$,$ b$,$ c$ be three positive real numbers such that $ab+bc+ca=3$ . For $\lambda\ge 2$ , prove that \[\frac{1}{(a+b)^2+\lambda}+\frac{1}{(b+c)^2+\lambda}+\frac{1}{(c+a)^2+\lambda}\le \frac{3}{4+\lambda}.\]Let $ a$,$ b$,$ c$ be three positive real numbers such that $ab+bc+ca=3$ . Prove that \[\frac{1}{4}+\frac{3}{8}\,{\frac {abc}{a+b+c}}\leq\frac{1}{(a+b)^2+4}+\frac{1}{(b+c)^2+4}+\frac{1}{(c+a)^2+4}\leq\frac{3}{8}.\]

Math-wiz wrote: We need to equivalently prove $$\sum\frac{(a+b)^2}{4+(a+b)^2}\geq\frac32$$By C-S, $$LHS\geq\frac{4(a+b+c)^2}{12+2(a^2+b^2+c^2+ab+bc+ca)}$$So it suffices to prove $8(a+b+c)^2\geq 36+6(a^2+b^2+c^2+ab+bc+ca)$ which is true by $a^2+b^2+c^2\geq ab+bc+ca$ pls explain the equivalent part

Professor33 wrote: Math-wiz wrote: We need to equivalently prove $$\sum\frac{(a+b)^2}{4+(a+b)^2}\geq\frac32$$By C-S, $$LHS\geq\frac{4(a+b+c)^2}{12+2(a^2+b^2+c^2+ab+bc+ca)}$$So it suffices to prove $8(a+b+c)^2\geq 36+6(a^2+b^2+c^2+ab+bc+ca)$ which is true by $a^2+b^2+c^2\geq ab+bc+ca$ pls explain the equivalent part $36=12(ab+bc+ca)$

MihaiT wrote: Professor33 wrote: Math-wiz wrote: We need to equivalently prove $$\sum\frac{(a+b)^2}{4+(a+b)^2}\geq\frac32$$By C-S, $$LHS\geq\frac{4(a+b+c)^2}{12+2(a^2+b^2+c^2+ab+bc+ca)}$$So it suffices to prove $8(a+b+c)^2\geq 36+6(a^2+b^2+c^2+ab+bc+ca)$ which is true by $a^2+b^2+c^2\geq ab+bc+ca$ pls explain the equivalent part $36=12(ab+bc+ca)$ bro in the starting when he wrote $$\sum\frac{(a+b)^2}{4+(a+b)^2}\geq\frac32$$PLS EXPLAIN THIS

I got the same solution as #2 Professor33 wrote: Math-wiz wrote: We need to equivalently prove $$\sum\frac{(a+b)^2}{4+(a+b)^2}\geq\frac32$$ pls explain the equivalent part