Áp dụng BĐT cosi: \(a+b\ge2\sqrt{ab}=2\cdot1=2\)

Vậy GTNN của a+b là 2, dấu \("="\Leftrightarrow a=b=1\) 

\(\hept{\begin{cases}a^2+b^2=t\\ab=m\end{cases}}\) \(t\ge0,m>0\)

\(\left(\sqrt{\frac{t}{m}}-\sqrt{\frac{m}{t}}\right)^2\ge0\)

\(\frac{t}{m}+\frac{m}{t}-2\sqrt{\frac{tm}{mt}}\ge0\)

\(\frac{t}{m}+\frac{m}{t}\ge2\sqrt{\frac{tm}{mt}=2}\)

min=2 , dấu = xảy ra khi \(\frac{t}{m}=\frac{m}{t}=1\)

\(\Leftrightarrow t=m\)

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

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

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

\(\left(a-b\right)^2+ab=0\)

\(\left(a-b\right)^2=-ab\) " vậy đề ngu =))

\(\left(a-b\right)^2=-ab\)

" tth m thông minh đấy nghĩ ra cái đề ? max óc 

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 \(a=b=\dfrac{1}{2}\)