Lời giải:
$\sqrt{ab}=1\Rightarrow ab=1$. Kết hợp với $b\geq 0\Rightarrow a>0$

Áp dụng BĐT Cô-si cho 2 số không âm:
$a+b\geq 2\sqrt{ab}=2$

Vậy $a+b_{\min}=2$. Giá trị này đạt tại $a=b>0$ và $ab=1$ hay $a=b=1$

Có: \(a^2+b^2\ge2ab\Rightarrow a^2+b^2\ge2\)
\(\Rightarrow\left(a+b+1\right)\left(a^2+b^2\right)\ge2\left(a+b+1\right)\)
\(\Rightarrow Q\ge2\left(a+b\right)+\frac{8}{a+b}+2\)
Mà: \(2\left(a+b\right)+\frac{8}{a+b}\ge2\sqrt{2\left(a+b\right).\frac{8}{a+b}}=8\)
\(\Rightarrow Q\ge10\)
Dấu "=" xảy ra <=> a=b=1

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 \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)

Đề bài nên là $a,b>0$ sao cho $a+b=1$

Lời giải:

Áp dụng BĐT  AM-GM:

$1=a+b\geq 2\sqrt{ab}\Rightarrow ab\leq \frac{1}{4}$
\(M=\frac{a^2+b^2}{ab}+ab=\frac{(a+b)^2-2ab}{ab}+ab=\frac{1}{ab}+ab-2\)

Tiếp tục áp dụng BĐT AM-GM:

\(ab+\frac{1}{16ab}\geq \frac{1}{2}\)

\(\frac{15}{16ab}\geq \frac{15}{16.\frac{1}{4}}=\frac{15}{4}\)

$\Rightarrow ab+\frac{1}{ab}\geq \frac{17}{4}$

$\Rightarrow M\geq \frac{9}{4}$

Vậy $M_{\min}=\frac{9}{4}$ khi $a=b=\frac{1}{2}$

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

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

\(S=\dfrac{1}{a^3+b^3}+\dfrac{1}{a^2b}+\dfrac{1}{ab^2}\ge\dfrac{1}{a^3+b^3}+\dfrac{4}{a^2b+ab^2}\)

\(S\ge\left(\dfrac{1}{a^3+b^3}+\dfrac{1}{a^2b+ab^2}+\dfrac{1}{a^2b+ab^2}+\dfrac{1}{a^2b+ab^2}\right)+\dfrac{1}{ab\left(a+b\right)}\)

\(S\ge\dfrac{16}{a^3+b^3+3a^2b+3ab^2}+\dfrac{1}{\dfrac{\left(a+b\right)^2}{4}.\left(a+b\right)}=\dfrac{20}{\left(a+b\right)^3}\ge20\)

\(S_{min}=20\) khi \(a=b=\dfrac{1}{2}\)

a)Có \(a^2+1\ge2a\) với mọi a; \(b^2+1\ge2b\) với mọi b

Cộng vế với vế \(\Rightarrow a^2+b^2+2\ge2\left(a+b\right)\)

Dấu = xảy ra <=> a=b=1

b) Áp dụng BĐT bunhiacopxki có:

\(\left(x+y\right)^2\le\left(1+1\right)\left(x^2+y^2\right)\Leftrightarrow\left(x+y\right)^2\le2\)

\(\Leftrightarrow-\sqrt{2}\le x+y\le\sqrt{2}\)

\(\Rightarrow\left(x+y\right)_{max}=\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x+y=\sqrt{2}\\x=y\end{matrix}\right.\)\(\Leftrightarrow x=y=\dfrac{\sqrt{2}}{2}\)

\(\left(x+y\right)_{min}=-\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x+y=-\sqrt{2}\\x=y\end{matrix}\right.\)\(\Leftrightarrow x=y=-\dfrac{\sqrt{2}}{2}\)

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

Với x,y>0, ta có: \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) (1)

Thật vậy (1) \(\Leftrightarrow\dfrac{y+x}{xy}\ge\dfrac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)\(\Leftrightarrow\left(x-y\right)^2\ge0\) (lđ)

Áp dụng (1) vào S ta được:

\(S\ge\dfrac{4}{a^2+b^2+2ab}+\dfrac{1}{2ab}\)

Lại có: \(ab\le\dfrac{\left(a+b\right)^2}{4}\) \(\Leftrightarrow2ab\le\dfrac{\left(a+b\right)^2}{2}\Leftrightarrow2ab\le\dfrac{1}{2}\)\(\Rightarrow\dfrac{1}{2ab}\ge2\)

\(\Rightarrow S\ge\dfrac{4}{\left(a+b\right)^2}+2=6\)

\(\Rightarrow S_{min}=6\Leftrightarrow a=b=\dfrac{1}{2}\)