\(P\ge\dfrac{\left(a+b\right)^2}{2ab}+\dfrac{\sqrt{ab}}{a+b}=\dfrac{\left(a+b\right)^2}{16ab}+\dfrac{\sqrt{ab}}{2\left(a+b\right)}+\dfrac{\sqrt{ab}}{2\left(a+b\right)}+\dfrac{7}{16}.\dfrac{\left(a+b\right)^2}{ab}\)

\(P\ge3\sqrt[3]{\dfrac{\left(a+b\right)^2ab}{64\left(a+b\right)^2.ab}}+\dfrac{7}{16}.\dfrac{4ab}{ab}=\dfrac{5}{2}\)

\(P_{min}=\dfrac{5}{2}\) khi \(a=b\)

Ta có: \(a^3 + b^3 = (a + b)^3 - 3ab(a + b)\)

\(a^3 + b^3 + 6ab \le 8\)

=>\((a + b)^3 - 3ab(a + b) + 6ab \le 8\)

\(\iff(a+b)^3-8-3ab(a+b-2)\le8\)

\(\iff(a+b-2)\left[(a+b)^2+2(a+b)+4\right]-3ab(a+b-2)\le0\)

\(\iff(a+b-2)\left[(a+b)^2+2(a+b)+4-3ab\right]\le0\)

Ta có: \((a+b)^2-3ab\)

\(=a^2-ab+b^2\)

\(=\left(a-\frac{b}{2}\right)^2+\frac{3b^2}{4}>0\quad(\text{ }a,b>0)\)

=>\((a+b)^2+2(a+b)+4-3ab>0\)

Do đó: a+b-2<=0

=>a+b<=2

Theo Cauchy, ta có: \(2 \ge a + b \ge 2\sqrt{ab} \implies \sqrt{ab} \le 1 \implies ab \le 1\)

\(P = \left( \frac{1}{a^2 + b^2} + \frac{1}{2ab} \right) + \frac{5}{2ab} + ab\)

Theo BĐT Cauchy-Schwarz, ta có:

\(\frac{1}{a^2 + b^2}+\frac{1}{2ab}\ge\frac{(1 + 1)^2}{a^2 + b^2 + 2ab}=\frac{4}{(a + b)^2}\ge\frac{4}{2^2}=1\)

\(\frac{5}{2ab} = \frac{1}{ab} + \frac{3}{2ab}\)

\(ab + \frac{1}{ab} \ge 2\sqrt{ab \cdot \frac{1}{ab}} = 2\)

mà ab<=1 nên \(\frac{3}{2ab} \ge \frac{3}{2 \cdot 1} = \frac{3}{2}\)

\(P = \left( \frac{1}{a^2 + b^2} + \frac{1}{2ab} \right) + \left( ab + \frac{1}{ab} \right) + \frac{3}{2ab}\)

=>\(P \ge 1 + 2 + \frac{3}{2} = \frac{9}{2}\)

Dấu '=' xảy ra khi a=b; a+b=2; ab=1

=>a=b=1

\(A=\dfrac{2}{a^2+b^2}+\dfrac{35}{ab}+2ab\\ =\dfrac{2}{a^2+b^2}+\dfrac{2}{2ab}+\dfrac{34}{ab}+\dfrac{17ab}{8}-\dfrac{ab}{8}\\ =2\left(\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}\right)+17\left(\dfrac{2}{ab}+\dfrac{ab}{8}\right)-\dfrac{ab}{8}\\ \overset{AM-GM}{\ge}2\cdot\dfrac{1}{a^2+b^2+2ab}+17\sqrt{\dfrac{2}{ab}\cdot\dfrac{ab}{8}}-\dfrac{\left(a+b\right)^2}{4\cdot8}\\ =\dfrac{2}{\left(a+b\right)^2}+\dfrac{17}{2}-\dfrac{\left(a+b\right)^2}{32}\\ \ge\dfrac{2}{4^2}+\dfrac{17}{2}-\dfrac{4^2}{32}=\dfrac{65}{8}\)

Dấu "=" xảy ra khi : \(\left\{{}\begin{matrix}\dfrac{2}{ab}=\dfrac{ab}{8}\\a^2+b^2=2ab\\a=b\\a+b=4\end{matrix}\right.\Leftrightarrow a=b=2\)

Vậy \(A_{Min}=\dfrac{65}{8}\) khi \(a=b=2\)

\(\ge2\cdot\dfrac{4}{a^2+b^2+2ab}+17\cdot2\sqrt{\dfrac{2}{ab}+\dfrac{ab}{8}}-\dfrac{\left(a+b\right)^2}{4\cdot8}\\ =\dfrac{8}{\left(a+b\right)^2}+17-\dfrac{\left(a+b\right)^2}{32}\\ \ge\dfrac{8}{4^2}+17-\dfrac{4^2}{32}=17\)

Vậy \(A_{Min}=17\) khi \(a=b=c=2\)

Áp dụng BĐT AM-GM ta có:

\(a+b\ge2\sqrt{ab}\Rightarrow1\ge2\sqrt{ab}\Rightarrow\dfrac{1}{2}\ge\sqrt{ab}\Rightarrow\dfrac{1}{4}\ge ab\)

Lại có theo AM-GM ta có:

\(a^2+b^2\ge2\sqrt{a^2b^2}=2ab\)\(\Rightarrow\dfrac{3}{a^2+b^2}\ge\dfrac{3}{2ab}\)

\(\Rightarrow A\ge\dfrac{3}{2ab}+\dfrac{2}{ab}\ge\dfrac{3}{2\cdot\dfrac{1}{4}}+\dfrac{2}{\dfrac{1}{4}}=14\)

Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}a+b=2\sqrt{ab}\\a+b=1\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=b\\a+b=1\end{matrix}\right.\)\(\Rightarrow a=b=\dfrac{1}{2}\)

Vậy \(A_{Min}=14\) khi \(a=b=\dfrac{1}{2}\)

\(A=\dfrac{3}{a^2+b^2}+\dfrac{3}{2ab}+\dfrac{1}{2ab}\ge\dfrac{12}{\left(a+b\right)^2}+\dfrac{2}{\left(a+b\right)^2}=14\)

\(C=\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{ab}+\dfrac{1}{ab}\right)+3\left(ab+\dfrac{1}{16ab}\right)+\dfrac{29}{16ab}\)

\(C\ge\dfrac{16}{a^2+b^2+2ab}+6\sqrt{\dfrac{ab}{16ab}}+\dfrac{29}{4\left(a+b\right)^2}\ge\dfrac{16}{1}+\dfrac{6}{4}+\dfrac{29}{4}=\dfrac{99}{4}\)

a: \(=\dfrac{\sqrt{a}-1}{\sqrt{a}\left(a-\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{1}\)

\(=a-1\)

b: \(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\left(\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}+\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}\right)\)

\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\dfrac{\sqrt{ab}+b+\sqrt{ab}-b}{\sqrt{a}\left(a-b\right)}\)

\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{1}{\sqrt{a}}\)

c: \(=\dfrac{a\sqrt{b}+b}{a-b}\cdot\sqrt{\dfrac{ab+b^2-2b\sqrt{ab}}{a^2+2a\sqrt{b}+b}}\cdot\left(\sqrt{a}+\sqrt{b}\right)\)

\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\sqrt{\dfrac{b\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(a+\sqrt{b}\right)^2}}\)

\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a+\sqrt{b}}=b\)

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 \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)