\(A=\dfrac{a+b+c}{a+\sqrt{\dfrac{a}{2}.2b}+\sqrt[3]{\dfrac{a}{4}.b.4c}}\ge\dfrac{a+b+c}{a+\dfrac{1}{2}\left(\dfrac{a}{2}+2b\right)+\dfrac{1}{3}\left(\dfrac{a}{4}+b+4c\right)}=\dfrac{3}{4}\)

\(A=\dfrac{1}{a^2+b^2}+\dfrac{1}{ab}+4ab=\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}+\dfrac{1}{2ab}+8ab-4ab\ge\dfrac{4}{a^2+b^2+2ab}+2\sqrt{\dfrac{1}{2}.8}-\dfrac{4.\left(a+b\right)^2}{4}=\dfrac{4}{\left(a+b\right)^2}+4-\left(a+b\right)^2\ge4+4-1=7\Rightarrow minA=7\Leftrightarrow a=b=\dfrac{1}{2}\)

\(M=\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\ge\dfrac{1}{a^2+b^2+c^2}+\dfrac{9}{ab+bc+ca}=\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{7}{ab+bc+ca}\)

\(M\ge\dfrac{9}{a^2+b^2+c^2+2ab+2bc+2ca}+\dfrac{7}{ab+bc+ca}=\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ca}\)

\(ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}\)

\(\Rightarrow M\ge\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ca}=9+\dfrac{7.3}{\left(a+b+c\right)^2}=9+21=30\)

\(Min_M=30\Leftrightarrow a=b=c=\dfrac{1}{3}\)

Áp dụng BĐT Svacxo

\(m\text{≥}\dfrac{1}{a^2+b^2+c^2}+\dfrac{9}{ab+bc+ca}\)

\(=\dfrac{1}{a^2+b^2+c^2}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{7}{ab+bc+ca}\)

≥ \(\dfrac{9}{a^2+b^2+c^2+2\left(ab+bc+ca\right)}\)\(+\dfrac{7}{ab+bc+ca}\)

\(=\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ca}\)

CM BĐT: \(a^2+b^2+c^2\text{≥}ab+bc+ca\)

⇔ \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\text{≥}0\) (luôn đúng)

⇒ \(\left(a+b+c\right)^2\text{≥}3\left(ab+bc+ca\right)\)

⇒ \(\dfrac{\left(a+b+c\right)^2}{3}\text{≥}ab+bc+ca\)

⇒ \(m\text{≥}\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{\dfrac{\left(a+b+c\right)^2}{3}}=9+21=30\) 

(vì a+b+c=1)

Vậy...

Ta có:

\(\dfrac{a}{bc}+\dfrac{b}{ca}\ge2\sqrt{\dfrac{ab}{abc^2}}=\dfrac{2}{c}\)

Tương tự: \(\dfrac{a}{bc}+\dfrac{c}{ab}\ge\dfrac{2}{b}\) ; \(\dfrac{b}{ca}+\dfrac{c}{ab}\ge\dfrac{2}{a}\)

Cộng vế với vế: \(\Rightarrow\dfrac{a}{bc}+\dfrac{b}{ca}+\dfrac{c}{ab}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)

\(\Rightarrow P\ge\dfrac{a^2+b^2+c^2}{2}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)

\(\Rightarrow P\ge\dfrac{1}{2}\left(a^2+\dfrac{1}{a}+\dfrac{1}{a}\right)+\dfrac{1}{2}\left(a^2+\dfrac{1}{b}+\dfrac{1}{b}\right)+\dfrac{1}{2}\left(c^2+\dfrac{1}{c}+\dfrac{1}{c}\right)\)

\(\Rightarrow P\ge\dfrac{1}{2}.3\sqrt[3]{\dfrac{a^2}{a^2}}+\dfrac{1}{2}.3\sqrt[3]{\dfrac{b^2}{b^2}}+\dfrac{1}{2}.3\sqrt[3]{\dfrac{c^2}{c^2}}=\dfrac{9}{2}\)

\(P_{min}=\dfrac{9}{2}\) khi \(a=b=c=1\)

\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)

\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)

\(\Leftrightarrow9abc\ge12\left(ab+bc+ca\right)-27\)

\(\Rightarrow abc\ge\dfrac{4}{3}\left(ab+bc+ca\right)-3\)

\(P\ge\dfrac{9}{a\left(b^2+bc+c^2\right)+b\left(c^2+ca+a^2\right)+c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}=\dfrac{9}{\left(ab+bc+ca\right)\left(a+b+c\right)}+\dfrac{abc}{ab+bc+ca}\)

\(\Rightarrow P\ge\dfrac{3}{ab+bc+ca}+\dfrac{abc}{ab+bc+ca}=\dfrac{3+abc}{ab+bc+ca}\)

\(\Rightarrow P\ge\dfrac{3+\dfrac{4}{3}\left(ab+bc+ca\right)-3}{ab+bc+ca}=\dfrac{4}{3}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

\(P=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}+\dfrac{1}{a^2+b^2+c^2}\ge\dfrac{\left(1+1+1\right)^2}{ab+bc+ca}+\dfrac{1}{a^2+b^2+c^2}\) (BĐT Cauchy Schwarz)

\(=\dfrac{9}{ab+bc+ca}+\dfrac{1}{a^2+b^2+c^2}\)

\(=\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{a^2+b^2+c^2}+\dfrac{7}{ab+bc+ca}\)

\(\ge\dfrac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2ab+2ac+2bc}+\dfrac{7}{ab+bc+ca}\)

\(=\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ca}\)

Ta có: \(ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{1}{3}\) .Thế vào biểu thức

\(\Rightarrow P\ge9+\dfrac{7}{\dfrac{1}{3}}=9+21=30\)

\(\Rightarrow P_{min}=30\) khi \(a=b=c=\dfrac{1}{3}\)