\(1,\text{Áp dụng Mincopxki: }\\ Q\ge\sqrt{\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2}\ge\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}\\ \text{Dấu }"="\Leftrightarrow a=b\)

\(2,\text{Áp dụng BĐT Cauchy-Schwarz: }\\ P\ge\dfrac{9}{a^2+b^2+c^2+2ab+2bc+2ca}=\dfrac{9}{\left(a+b+c\right)^2}\ge\dfrac{9}{1}=9\\ \text{Dấu }"="\Leftrightarrow a=b=c=\dfrac{1}{3}\)

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

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

Tương tự

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

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

2 ) Ta có : \(\frac{1}{3}\left(a^3+b^3+a+b\right)+ab\le a^2+b^2+1\)

\(\Leftrightarrow\frac{1}{3}\left(a+b\right)\left(a^2+b^2+1-ab\right)+ab\le a^2+b^2+1\)

\(\Leftrightarrow\left(a^2+b^2+1\right)\left(\frac{a+b}{3}-1\right)-ab\left(\frac{a+b}{3}-1\right)\le0\)

\(\Leftrightarrow\left(a^2+b^2+1-ab\right)\left(\frac{a+b}{3}-1\right)\le0\)

Do a ; b dương \(\Rightarrow a^2+b^2+1-ab>0\Rightarrow\frac{a+b}{3}-1\le0\)

\(\Leftrightarrow a+b\le3\)

\(M=\frac{a^2+8}{a}+\frac{b^2+2}{b}=a+\frac{8}{a}+b+\frac{2}{b}=2a+\frac{8}{a}+\frac{2}{b}+2b-\left(a+b\right)\ge8+4-3=9\)

( áp dụng BĐT Cauchy cho a ; b dương )

Dấu " = " xảy ra \(\Leftrightarrow a=2;b=1\)

Tìm min cho K, tìm max có lẽ Bunhia là ra thôi:

Đặt \(\left\{{}\begin{matrix}\sqrt{3a+1}=x\\\sqrt{3b+1}=y\\\sqrt{3x+1}=z\end{matrix}\right.\) \(\Rightarrow1\le x;y;z\le\sqrt{10}\)

\(x^2+y^2+z^2=3\left(a+b+c\right)+3=12\)

Bài toán trở thành cho \(x^2+y^2+z^2=12\), tìm min \(P=x+y+z\)

Ta có: \(\left(x-1\right)\left(x-\sqrt{10}\right)\le0\Rightarrow x^2-\left(\sqrt{10}+1\right)x+\sqrt{10}\le0\)

\(\left(y-1\right)\left(y-\sqrt{10}\right)=y^2-\left(\sqrt{10}+1\right)y+\sqrt{10}\le0\)

\(\left(z-1\right)\left(z-\sqrt{10}\right)=z^2-\left(\sqrt{10}+1\right)z+\sqrt{10}\le0\)

Cộng vế với vế:

\(x^2+y^2+z^2-\left(\sqrt{10}+1\right)\left(x+y+z\right)+3\sqrt{10}\le0\)

\(\Rightarrow x+y+z\ge\frac{x^2+y^2+z^2+3\sqrt{10}}{\sqrt{10}+1}=\frac{12+3\sqrt{10}}{\sqrt{10}+1}=2+\sqrt{10}\)

\(\Rightarrow P_{min}=2+\sqrt{10}\) khi \(\left(x;y;z\right)=\left(1;1;\sqrt{10}\right)\) và các hoán vị hay \(\left(a;b;c\right)=\left(3;0;0\right)\) và các hoán vị

haha

\(\sqrt[3]{a^2+\frac{1}{a^2}}+\sqrt[3]{b^2+\frac{1}{b^2}}\ge2\sqrt[6]{\frac{a^2}{b^2}+\frac{b^2}{a^2}+a^2b^2+\frac{1}{a^2b^2}}\ge2\sqrt[6]{2+a^2b^2+\frac{1}{a^2b^2}}\)

đến đây thì ta thấy từ giả thuyết có \(a+b=\frac{2}{3}\Rightarrow a^2b^2\le\frac{1}{81}\)

Xét:

\(a^2b^2+\frac{1}{a^2b^2}=\left(a^2b^2+\frac{1}{6561a^2b^2}\right)+\frac{6560}{6561a^2b^2}\ge\frac{6562}{81}\)

\(\Rightarrow\sqrt[3]{a^2+\frac{1}{a^2}}+\sqrt[3]{b^2+\frac{1}{b^2}}\ge2\sqrt[3]{\frac{82}{9}}\)

Đẳng thức xảy ra khi \(a=b=\frac{1}{3}\)

`1. P = x/(sqrt x-1)`

`= (x-1+1)/(sqrtx-1)`

`= ((sqrt x+1)(sqrt x-1))/(sqrt x-1) +1/(sqrt x-1)`

`= sqrt x+1 + 1/(sqrt x-1)`

`= sqrtx-1 + 1/(sqrt x-1) + 2 >= 4`.

ĐTXR `<=> (sqrtx-1)^2 = 1`.

`<=> x =4` hoặc `x = 0 ( ktm)`.

Vậy Min A `= 4 <=> x= 4`.

1) \(P=\dfrac{x}{\sqrt{x}-1}=\dfrac{(x-\sqrt{x})+(\sqrt{x}-1)+1}{\sqrt{x}-1}=\sqrt{x}+\dfrac{1}{\sqrt{x}-1}+1\)

\(=\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+2\)

Với x>1\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x}-1>0\\\dfrac{1}{\sqrt{x}-1}>0\end{matrix}\right.\)

Áp dụng BĐT AM-GM cho 2 số dương \(\sqrt{x}-1\) và \(\dfrac{1}{\sqrt{x}-1}\), ta có:

\(\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}\ge2\sqrt{(\sqrt{x}-1).\dfrac{1}{\sqrt{x}-1}}=2\)

\(\Rightarrow P\ge2+2=4\)

Dấu = xảy ra khi: \(\sqrt{x}-1=1\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)

KL;....

2:

\(B=\dfrac{x+16}{\sqrt{x}+3}=\dfrac{x-9+25}{\sqrt{x}+3}\)

\(=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6\)

=>\(B>=2\cdot\sqrt{25}-6=4\)

Dấu = xảy ra khi (căn x+3)^2=25

=>căn x+3=5

=>căn x=2

=>x=4