\(1,Q=\dfrac{a^4-2a^2+a^3-2a+a^2-2}{a^4-2a^2+2a^3-4a+a^2-2}\\ Q=\dfrac{\left(a^2-2\right)\left(a^2+a+1\right)}{\left(a^2-2\right)\left(a^2+2a+1\right)}=\dfrac{a^2+a+1}{a^2+2a+1}\)

\(Q=\dfrac{x^2+x+1}{\left(x+1\right)^2}-\dfrac{3}{4}+\dfrac{3}{4}=\dfrac{x^2+x+1-\dfrac{3}{4}x^2-\dfrac{3}{2}x-\dfrac{3}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}\\ Q=\dfrac{\dfrac{1}{4}x^2-\dfrac{1}{2}x+\dfrac{1}{4}}{\left(x+1\right)^2}+\dfrac{3}{4}=\dfrac{\dfrac{1}{4}\left(x-1\right)^2}{\left(x+1\right)^2}+\dfrac{3}{4}\ge\dfrac{3}{4}\\ Q_{min}=\dfrac{3}{4}\Leftrightarrow x=1\)

\(2,\text{Từ GT }\Leftrightarrow\dfrac{ayz+bxz+czy}{xyz}=0\\ \Leftrightarrow ayz+bxz+czy=0\\ \text{Ta có }\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\\ \Leftrightarrow\left(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\right)^2=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\left(\dfrac{xy}{ab}+\dfrac{yz}{bc}+\dfrac{zx}{ca}\right)=0\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{cxy+ayz+bzx}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}+2\cdot\dfrac{0}{abc}=1\\ \Leftrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1\)

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 \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)

áp dụng BDT AM-GM \(=>x+y\ge2\sqrt{xy}=>\left(x+y\right)^2\ge4xy\left(1\right)\)

mà \(x+y\le1=>\left(x+y\right)^2\le1\left(2\right)\)

(1)(2)\(=>4xy\le\left(x+y\right)^2\le1=>4xy\le1=>xy\le\dfrac{1}{4}\)

\(A=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\sqrt{1+x^2y^2}\ge2\sqrt{\dfrac{1+x^2y^2}{xy}}=2\sqrt{\dfrac{1}{xy}+xy}\)

\(=2\sqrt{\dfrac{1}{xy}+16xy-15xy}=2\sqrt{2\sqrt{16}-\dfrac{15}{4}}=\sqrt{17}\)

dấu"=" xảy ra<=>\(x=y=\dfrac{1}{2}\)

\(1\ge x+y\ge2\sqrt{xy}\Rightarrow xy\le\dfrac{1}{4}\Rightarrow\dfrac{1}{xy}\ge4\)

Ta có:

\(A\ge\dfrac{2}{\sqrt{xy}}.\sqrt{1+x^2y^2}=2\sqrt{\dfrac{1}{xy}+xy}=2\sqrt{\left(xy+\dfrac{1}{16xy}\right)+\dfrac{15}{16}.\dfrac{1}{xy}}\)

\(A\ge2\sqrt{2\sqrt{\dfrac{xy}{16xy}}+\dfrac{15}{16}.4}=\sqrt{17}\)

\(A_{min}=\sqrt{17}\) khi \(x=y=\dfrac{1}{2}\)

kẻ lười biếng nạp card, đi ô tô

\(\dfrac{\left(x+y+1\right)^2}{xy+x+y}\ge\dfrac{3\left(xy+x+y\right)}{xy+x+y}=3\)

\(\Rightarrow A=\dfrac{8\left(x+y+1\right)^2}{9\left(xy+x+y\right)}+\dfrac{\left(x+y+1\right)^2}{9\left(xy+x+y\right)}+\dfrac{xy+x+y}{\left(x+y+1\right)^2}\)

\(A\ge\dfrac{8}{9}.3+2\sqrt{\dfrac{\left(x+y+1\right)^2\left(xy+x+y\right)}{\left(xy+x+y\right)\left(x+y+1\right)^2}}=\dfrac{10}{3}\)

Dấu "=" xảy ra khi \(x=y=1\)

Lời giải:

$A=(x-y)+\frac{4}{x-y}+y+\frac{1}{y}$

Áp dụng BĐT Cô-si:

$(x-y)+\frac{4}{x-y}\geq 2\sqrt{(x-y).\frac{4}{x-y}}=4$
$y+\frac{1}{y}\geq 2$

$\Rightarrow A\geq 4+2=6$

Vậy $A_{\min}=6$ khi $(x,y)=(3,1)$

ÁP dụng BĐT Mincopxki, ta có:

\(A\ge\sqrt{\left(x+y\right)^2+\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2}\)

\(=\sqrt{\left(x+y\right)^2+\dfrac{\left(x+y\right)^2}{\left(xy\right)^2}}\)

\(\ge\sqrt{2\sqrt{\left(x+y\right)^2.\dfrac{\left(x+y\right)^2}{\left(xy\right)^2}}}=\sqrt{\dfrac{2\left(x+y\right)^2}{xy}}\) (cô si)

\(\ge\sqrt{\dfrac{2.4xy}{xy}}=\sqrt{8}=2\sqrt{2}\left(Côsi\right)\)

Min \(A=2\sqrt{2}\Leftrightarrow x=y\)

Ta có : \(\dfrac{a+b}{2}\ge\sqrt{ab}\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\) ( luôn đúng )

Dấu "=" xảy ra khi \(a=b\)

Bài tập :

Có : \(A=\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{x+y}{x}+\dfrac{x+y}{y}=2+\dfrac{x}{y}+\dfrac{y}{x}\) ( do \(x+y=1\) )

Theo BĐT trên có : \(\dfrac{x}{y}+\dfrac{y}{x}\ge2.\sqrt{\dfrac{x}{y}\cdot\dfrac{y}{x}}=2\)

Nên \(A\ge2+2=4\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)