Let $a,b,c\ge 0: a+b+c=3.$ Prove that $$\frac{a}{a^2+3}+\frac{b}{b^2+3}+\frac{c}{c^2+3}\le \frac{ab+bc+ca+3}{8}$$ I'm looking for a smooth proof by using classical inequalities as AM-GM, Cauchy-Schwarz,... In my proof, I used $$\frac{a}{a^2+3}\le \frac{4a-a^2+1}{16}.$$ The result follows.
Notice: $$\frac{16a}{a^2+3}+a^2-1-4a=-\frac{(3-a)(a+1)(a-1)^2}{a^2+3}\le 0.$$ Thank you.
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By C-S $$\frac{ab+ac+bc+3}{8}-\sum_{cyc}\frac{a}{a^2+3}=\frac{ab+ac+bc+3}{8}-\sum_{cyc}\left(\frac{a}{a^2+3}-\frac{a}{3}\right)-1=$$ $$\frac{ab+ac+bc-5}{8}+\sum_{cyc}\frac{a^3}{3(a^2+3}\geq\frac{ab+ac+bc-5}{8}+\frac{(a^2+b^2+c^2)^2}{3\sum\limits_{cyc}(a^3+3a)}=$$ $$=\frac{9(ab+ac+bc)-5(a+b+c)^2}{24(a+b+c)}+\frac{(a^2+b^2+c^2)^2}{3(a^3+b^3+c^3)+(a+b+c)^3)}\geq0$$ because the last inequality it's $$\sum_{cyc}(4a^5+5a^4b+5a^4c+10a^3b^2+10a^3c^2-40a^3bc+6a^2b^2c)\geq0,$$ which is true by AM-GM and Muthead: $$\sum_{cyc}(4a^5+5a^4b+5a^4c+10a^3b^2+10a^3c^2-40a^3bc+6a^2b^2c)=$$ $$=\sum_{cyc}(4a^5+5a^4b+5a^4c+10a^3b^2+10a^3c^2+3a^2b^2c+3a^2c^2b-40a^3bc)\geq$$ $$\geq\sum_{cyc}\left(40\sqrt[40]{a^{20+20+20+30+30+6+6}b^{5+20+6+3}c^{5+20+3+6}}-40a^3bc\right)=$$ $$=40\sum_{cyc}\left(a^{3.3}b^{0.85}c^{0.85}-a^3bc\right)\geq0$$
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Alternative proof.
By using Cauchy-Schwarz $$a^2+3=a^2+a+b+c=a^2+\frac{1}{\dfrac{1}{a}}+\frac{b^2}{b}+\frac{c^2}{c}\ge \frac{(a+b+c+1)^2}{1+\dfrac{1}{a}+b+c}.$$It implies that $$\frac{1}{a^2+3}\le \frac{1+\dfrac{1}{a}+b+c}{16},$$or$$\frac{a}{a^2+3}\le \frac{a+1+ab+ac}{16}.$$ Similarly, we obtain$$\frac{a}{a^2+3}+\frac{b}{b^2+3}+\frac{c}{c^2+3}\le \frac{2(ab+bc+ca)+6}{16}=\frac{ab+bc+ca+3}{8}.$$ The proof is done. Equality holds at $a=b=c=1.$
It is quite similar as yours UCT but more smooth, I think.
$$\sum_{cyc}\frac{a}{a^2+3}\le \frac{4(a+b+c)-(a^2+b^2+c^2)+3}{16} $$
and we know that :$a+b+c=3$ $$\implies a^2+b^2+c^2 = 9-2(ab+bc+ac)$$
Finally : $$\sum_{cyc}\frac{a}{a^2+3}\le \frac{4(a+b+c)-(a^2+b^2+c^2)+3}{16} =\frac{6+2(ab+bc+ac)}{16} =\frac{3+(ab+bc+ac)}{8}$$