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Cách 1:$AM-GM:\sqrt{2b(a+b)}\leq \frac{a+3b}{2}$Do đó: $P\geq \Sigma \frac{2a\sqrt{2}}{a+3b} $$\rightarrow $ Ta chứng minh: $S=\Sigma \frac{a}{a+3b}\geq \frac{3}{4}$hay $S=\Sigma \frac{a^2}{a^2+3ab}\geq \frac{3}{4}$$ Cauchy-Schwarz:$$S\geq \frac{(a+b+c)^2}{a^2+b^2+c^2+3ab+3bc+3ca}=\frac{(a+b+c)^2}{a^2+b^2+c^2+\frac{8}{3}(ab+bc+ca)+\frac{1}{3}(ab+bc+ca)}\geq \frac{(a+b+c)^2}{\frac{4}{3}(a^2+b^2+c^2)+\frac{8}{3}(ab+bc+ca)}=\frac{(a+b+c)^2}{\frac{4}{3}(a+b+c)^2}=\frac{3}{4}$Do đó: $P\geq 2\sqrt{2}.\frac{3}{4}=\frac{3\sqrt{2}}{2}$Đẳng thức khi$a=b=c./$

Cách 1:$AM-GM:\sqrt{2b(a+b)}\leq \frac{a+3b}{2}$Do đó: $A\geq \Sigma \frac{2a\sqrt{2}}{a+3b} $$\rightarrow $ Ta chứng minh: $S=\Sigma \frac{a}{a+3b}\geq \frac{3}{4}$hay $S=\Sigma \frac{a^2}{a^2+3ab}\geq \frac{3}{4}$$ Cauchy-Schwarz:$$S\geq \frac{(a+b+c)^2}{a^2+b^2+c^2+3ab+3bc+3ca}=\frac{(a+b+c)^2}{a^2+b^2+c^2+\frac{8}{3}(ab+bc+ca)+\frac{1}{3}(ab+bc+ca)}\geq \frac{(a+b+c)^2}{\frac{4}{3}(a^2+b^2+c^2)+\frac{8}{3}(ab+bc+ca)}=\frac{(a+b+c)^2}{\frac{4}{3}(a+b+c)^2}=\frac{3}{4}$Do đó: $A\geq 2\sqrt{2}.\frac{3}{4}=\frac{3\sqrt{2}}{2}$Đẳng thức khi$a=b=c./$