1) Áp dụng bđt Cauchy cho 3 số dương ta có

 \(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{x}+x^3\ge4\sqrt[4]{\dfrac{1}{x}.\dfrac{1}{x}.\dfrac{1}{x}.x^3}=4\) (1)

\(\dfrac{3}{y^2}+y^2\ge2\sqrt{\dfrac{3}{y^2}.y^2}=2\sqrt{3}\) (2)

\(\dfrac{3}{z^3}+z=\dfrac{3}{z^3}+\dfrac{z}{3}+\dfrac{z}{3}+\dfrac{z}{3}\ge4\sqrt[4]{\dfrac{3}{z^3}.\dfrac{z}{3}.\dfrac{z}{3}.\dfrac{z}{3}}=4\sqrt{3}\) (3)

Cộng (1);(2);(3) theo vế ta được

\(\left(\dfrac{3}{x}+\dfrac{3}{y^2}+\dfrac{3}{z^3}\right)+\left(x^3+y^2+z\right)\ge4+2\sqrt{3}+4\sqrt{3}\)

\(\Leftrightarrow3\left(\dfrac{1}{x}+\dfrac{1}{y^2}+\dfrac{1}{z^3}\right)\ge3+4\sqrt{3}\)

\(\Leftrightarrow P\ge\dfrac{3+4\sqrt{3}}{3}\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{x}=x^3\\\dfrac{3}{y^2}=y^2\\\dfrac{3}{z^3}=\dfrac{z}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\sqrt[4]{3}\\z=\sqrt{3}\end{matrix}\right.\) (thỏa mãn giả thiết ban đầu)

2) Ta có \(4\sqrt{ab}=2.\sqrt{a}.2\sqrt{b}\le a+4b\)

Dấu"=" khi a = 4b

nên \(\dfrac{8}{7a+4b+4\sqrt{ab}}\ge\dfrac{8}{7a+4b+a+4b}=\dfrac{1}{a+b}\)

Khi đó \(P\ge\dfrac{1}{a+b}-\dfrac{1}{\sqrt{a+b}}+\sqrt{a+b}\)

Đặt \(\sqrt{a+b}=t>0\) ta được

\(P\ge\dfrac{1}{t^2}-\dfrac{1}{t}+t=\left(\dfrac{1}{t^2}-\dfrac{2}{t}+1\right)+\dfrac{1}{t}+t-1\)

\(=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\)

Có \(\dfrac{1}{t}+t\ge2\sqrt{\dfrac{1}{t}.t}=2\) (BĐT Cauchy cho 2 số dương)

nên \(P=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\ge\left(\dfrac{1}{t}-1\right)^2+1\ge1\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{t}-1=0\\t=\dfrac{1}{t}\end{matrix}\right.\Leftrightarrow t=1\)(tm)

khi đó a + b = 1

mà a = 4b nên \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)

Vậy MinP = 1 khi \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)

\(S=\sqrt{a^2+\dfrac{1}{b^2}}+\sqrt{b^2+\dfrac{1}{c^2}}+\sqrt{c^2+\dfrac{1}{a^2}}\)

\(\sqrt{a^2+\dfrac{1}{b^2}}=\dfrac{1}{\sqrt{17}}\sqrt{\left(a^2+\dfrac{1}{b^2}\right)\left(1+4^2\right)}\ge\dfrac{1}{\sqrt{17}}\left(a+\dfrac{4}{b}\right)\left(1\right)\)\(\left(bunhia\right)\)

\(tương-tự\Rightarrow\sqrt{b^2+\dfrac{1}{c^2}}\ge\dfrac{1}{\sqrt{17}}\left(b+\dfrac{4}{c}\right)\left(2\right)\)

\(\sqrt{c^2+\dfrac{1}{a^2}}\ge\dfrac{1}{\sqrt{17}}\left(c+\dfrac{4}{a}\right)\left(3\right)\)

\(\left(1\right)\left(2\right)\left(3\right)\Rightarrow S\ge\dfrac{1}{\sqrt{17}}\left(a+\dfrac{4}{b}+b+\dfrac{4}{c}+c+\dfrac{4}{a}\right)\)

\(\Rightarrow S\ge\dfrac{1}{\sqrt{17}}\left[16a+\dfrac{4}{a}+16b+\dfrac{4}{b}+16c+\dfrac{4}{c}-15\left(a+b+c\right)\right]\)

\(\Rightarrow S\ge\dfrac{1}{\sqrt{17}}\left[2\sqrt{16a.\dfrac{4}{a}}+2\sqrt{16b.\dfrac{4}{b}}+2\sqrt{16c.\dfrac{4}{c}}-15.\dfrac{3}{2}\right]\left(am-gm\right)\)

\(\Rightarrow S\ge\dfrac{1}{\sqrt{17}}\left(16+16+16-\dfrac{45}{2}\right)=\dfrac{3\sqrt{17}}{2}\)

\(\Rightarrow MinS=\dfrac{3\sqrt{17}}{2}\Leftrightarrow a=b=c=\dfrac{1}{2}\)

\(ab\cdot\sqrt{\dfrac{a}{3b}}-a^2\sqrt{\dfrac{3b}{a}}\)

\(=a\sqrt{ab}-a^2\cdot\dfrac{\sqrt{3b}}{\sqrt{a}}\)

\(=a\sqrt{ab}-a\sqrt{a}\cdot\sqrt{3b}\)

\(=a\sqrt{ab}\left(1-\sqrt{3}\right)\)

\(\Leftrightarrow m=\dfrac{a\sqrt{ab}\left(1-\sqrt{3}\right)}{\sqrt{3ab}}=\dfrac{a\left(\sqrt{3}-3\right)}{3}\)

\(M\ge\dfrac{\sqrt{\left(\sqrt{a}+\sqrt{b}\right)^2}}{2}+\dfrac{\sqrt{\left(\sqrt{b}+\sqrt{c}\right)^2}}{2}+\dfrac{\sqrt{\left(\sqrt{c}+\sqrt{a}\right)^2}}{2}\)

\(M\ge\sqrt{a}+\sqrt{b}+\sqrt{c}=3\)

Dấu "=" xảy ra khi \(a=b=c=1\)

\(=\left(1^2+4^2\right)\left(a^2+\dfrac{1}{b^2}\right)\ge\left(1a+4.\dfrac{1}{b}\right)^2\\ \Rightarrow\sqrt{a^2+\dfrac{1}{vb^2}}\ge\dfrac{1}{\sqrt{17}}\left(a+\dfrac{4}{b}\right)\) 

Tương tự

\(\sqrt{b^2+\dfrac{1}{c^2}}\ge\dfrac{1}{\sqrt{17}}\left(b+\dfrac{4}{c}\right)\\ \sqrt{c^2+\dfrac{1}{a^2}}\ge\dfrac{1}{\sqrt{17}}\left(c+\dfrac{4}{a}\right)\\ Do.đó:\\ Q\ge\dfrac{1}{\sqrt{17}}\left(a+b+c+\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)\ge\dfrac{1}{\sqrt{17}}\\ \left(a+b+c+\dfrac{36}{a+b+c}\right)\) 

\(=\dfrac{1}{\sqrt{17}}\\ \left[a+b+c+\dfrac{9}{4\left(a+b+c\right)}+\dfrac{135}{4\left(a+b+c\right)}\right]\\ \ge\dfrac{3\sqrt{17}}{2}\)

\(\dfrac{a}{\sqrt{a^2+15bc}}+\dfrac{b}{\sqrt{b^2+15ca}}+\dfrac{c}{\sqrt{c^2+15ab}}\ge\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{a^2}{a\sqrt{a^2+15bc}}+\dfrac{b^2}{b\sqrt{b^2+15ca}}+\dfrac{c^2}{c\sqrt{c^2+15ab}}\ge\dfrac{3}{4}\)

Áp dụng BĐT Caushy-Schwarz ta được:

\(\dfrac{a^2}{a\sqrt{a^2+15bc}}+\dfrac{b^2}{b\sqrt{b^2+15ca}}+\dfrac{c^2}{c\sqrt{c^2+15ab}}\ge\dfrac{\left(a+b+c\right)^2}{a\sqrt{a^2+15bc}+b\sqrt{b^2+15ca}+c\sqrt{c^2+15ab}}\)

Ta chứng minh rằng:

\(a\sqrt{a^2+15bc}+b\sqrt{b^2+15ca}+c\sqrt{c^2+15ab}\le\dfrac{4}{3}\left(a+b+c\right)^2\)

\(\Leftrightarrow\sqrt{a}\sqrt{a^3+15abc}+\sqrt{b}\sqrt{b^3+15abc}+\sqrt{c}\sqrt{c^3+15abc}\le\dfrac{4}{3}\left(a+b+c\right)^2\)

Áp dụng BĐT Bunhiacopxki ta được:

\(\sqrt{a}\sqrt{a^3+15abc}+\sqrt{b}\sqrt{b^3+15abc}+\sqrt{c}\sqrt{c^3+15abc}\le\sqrt{\left(a+b+c\right)\left(a^3+b^3+c^3+45abc\right)}\)Ta tiếp tục chứng minh:

\(\dfrac{16}{9}\left(a+b+c\right)^3\ge a^3+b^3+c^3+45abc\)

\(\Leftrightarrow\dfrac{16}{9}\left(a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\right)\ge a^3+b^3+c^3+45abc\)

Áp dụng BĐT AM-GM (Caushy) ta được:

\(\dfrac{16}{9}\left(a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\right)\ge\dfrac{16}{9}\left(a^3+b^3+c^3+3.2\sqrt{ab}.2.\sqrt{bc}.2.\sqrt{ca}\right)=\dfrac{16}{9}.\left(a^3+b^3+c^3+24abc\right)\)

Ta chứng minh:

\(\dfrac{16}{9}\left(a^3+b^3+c^3+24abc\right)\ge a^3+b^3+c^3+45abc\)

\(\Leftrightarrow\dfrac{16}{9}a^3+\dfrac{16}{9}b^3+\dfrac{16}{9}c^3+\dfrac{16}{9}.24abc\ge a^3+b^3+c^3+45abc\)

\(\Leftrightarrow\dfrac{7}{9}\left(a^3+b^3+c^3\right)\ge\dfrac{7}{3}abc\) (*)

Áp dụng BĐT AM-GM (Caushy) ta được:

\(\dfrac{7}{9}\left(a^3+b^3+c^3\right)\ge\dfrac{7}{9}.3\sqrt[3]{a^3b^3c^3}=\dfrac{7}{3}abc\)

\(\Rightarrow\) (*) đúng.

Vậy BĐT đã được chứng minh. Dấu "=" xảy ra khi \(a=b=c>0\).