Áp dụng Bất Đẳng Thức Cosi ta có \(\hept{\begin{cases}\frac{x^3}{1+y}+\frac{1+y}{4}+\frac{1}{2}\ge3\sqrt[3]{\frac{x^3}{1+y}\cdot\frac{1+y}{4}\cdot\frac{1}{2}}=\frac{3x}{2}\\\frac{y^3}{1+z}+\frac{1+z}{4}+\frac{1}{2}\ge3\sqrt[3]{\frac{y^3}{1+z}\cdot\frac{1+z}{4}\cdot\frac{1}{2}}=\frac{3y}{2}\\\frac{z^3}{1+x}+\frac{1+x}{4}+\frac{1}{2}\ge3\sqrt[3]{\frac{z^3}{1+x}\cdot\frac{1+x}{4}\cdot\frac{1}{2}}=\frac{3z}{2}\end{cases}}\)

Cộng vế theo vế ta được \(P+\frac{3+x+y+z}{4}+\frac{3}{2}\ge\frac{3}{2}\left(x+y+z\right)\)

\(\Leftrightarrow P\ge\frac{5}{4}\left(x+y+z\right)-\frac{9}{4}\)

Mà ta có \(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\ge9\Rightarrow x+y+z\ge3\)

Do đó \(P\ge\frac{5}{4}\cdot3-\frac{9}{4}=\frac{3}{2}\). Dấu "=" xảy ra khi x=y=z=1

Vậy minP=\(\frac{3}{2}\)khi x=y=z=1

\(T=x^2+y^2+\frac{1}{x}+\frac{1}{x+y}\)

\(=\left(x-2\right)^2+\left(y-1\right)^2+\left(\frac{x}{4}+\frac{1}{x}\right)+\left(\frac{x+y}{9}+\frac{1}{x+y}\right)+\frac{17}{9}\left(x+y\right)+\frac{7x}{9}-5\)

\(\ge0+0+2\sqrt{\frac{x}{4}\cdot\frac{1}{x}}+2\sqrt{\frac{x+y}{9}\cdot\frac{1}{x+y}}+\frac{17\cdot3}{9}+\frac{7\cdot2}{9}-5\)

\(=\frac{35}{9}\)

Đẳng thức xảy ra tại x=2;y=1

Đặt x = 2t 

đưa bài toán về dạng: 

\(T=4t^2+y^2+\frac{1}{2t}+\frac{1}{2t+y}\ge\left(t^2+t^2+y^2\right)+\frac{1}{2t+y}+\left(2t^2+\frac{1}{2t}\right)\)

\(\ge\frac{\left(2t+y\right)^2}{3}+\frac{1}{2t+y}+\left(2t^2+\frac{1}{2t}\right)\)

\(=\left(\frac{\left(2t+y\right)^2}{3}+\frac{9}{2t+y}+\frac{9}{2t+y}\right)+\left(2t^2+\frac{4}{2t}+\frac{4}{2t}\right)-\frac{17}{2t+y}-\frac{7}{2t}\)

\(\ge3.3+3.2-\frac{17}{3}-\frac{7}{2}=\frac{35}{6}\)

Dấu "=" xảy ra <=> y = t = 1 <=> y = 1 ; x = 2

Dòng 2 là \(\frac{7x}{4}\)

\(A=3\left(\frac{\sqrt{x-3}-\sqrt{x}+\sqrt{x-3}+\sqrt{x}}{\left(\sqrt{x-3}+\sqrt{x}\right)\left(\sqrt{x-3}-\sqrt{x}\right)}\right)+\frac{x\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)

\(=3\left(\frac{2\sqrt{x-3}}{-3}\right)+x=x-2\sqrt{x-3}\)

\(A=x-3-2\sqrt{x-3}+1+2=\left(\sqrt{x-3}-1\right)^2+2\ge2\)

\(A_{min}=2\) khi \(\sqrt{x-3}=1\Leftrightarrow x=4\)

Cách khác:

Áp dụng BĐT Cauchy-Schwarz:

\(P=\frac{x^4}{x+xy}+\frac{y^4}{y+yz}+\frac{z^4}{z+zx}\geq \frac{(x^2+y^2+z^2)^2}{x+y+z+xy+yz+xz}\)

Áp dụng BĐT AM-GM ta có:

\(x^2+y^2+z^2\geq xy+yz+xz(1)\)

\(\Rightarrow 2(x^2+y^2+z^2)\geq 2(xy+yz+xz)\)

\(\Rightarrow 3(x^2+y^2+z^2)\geq (x+y+z)^2\)

\(\Rightarrow (x+y+z)^2\leq 3(x^2+y^2+z^2)\leq (xy+yz+xz)(x^2+y^2+z^2)\leq (x^2+y^2+z^2)^2\)

\(\Rightarrow x+y+z\le x^2+y^2+z^2(2)\)

Từ $(1);(2)$ suy ra:

\(P\geq \frac{(x^2+y^2+z^2)^2}{2(x^2+y^2+z^2)}=\frac{x^2+y^2+z^2}{2}\geq \frac{xy+yz+xz}{2}\geq \frac{3}{2}\)

Vậy $P_{\min}=\frac{3}{2}$

Lời giải:

Áp dụng BĐT AM-GM:

\(\frac{x^3}{y+1}+\frac{y+1}{4}+\frac{1}{2}\geq 3\sqrt[3]{\frac{x^3}{y+1}.\frac{y+1}{4}.\frac{1}{2}}=\frac{3x}{2}\)

\(\frac{y^3}{z+1}+\frac{z+1}{4}+\frac{1}{2}\geq \frac{3y}{2}\)

\(\frac{z^3}{1+x}+\frac{1+x}{4}+\frac{1}{2}\geq \frac{3z}{2}\)

Cộng theo vế và thu gọn:

\(\Rightarrow P\geq \frac{5}{4}(x+y+z)-\frac{9}{4}\)

Theo hệ quả quen thuộc của BĐT AM-GM:

\((x+y+z)^2\geq 3(xy+yz+xz)\geq 9\)

\(\Rightarrow x+y+z\geq 3\)

\(\Rightarrow P\geq \frac{5}{4}(x+y+z)-\frac{9}{4}\geq \frac{5}{4}.3-\frac{9}{4}=\frac{3}{2}\)

Vậy $P_{\min}=\frac{3}{2}$ khi $x=y=z=1$

Lời giải:
1. Áp dụng BĐT Cô-si

$G=\frac{x^2}{x-1}=\frac{(x^2-1)+1}{x-1}=x+1+\frac{1}{x-1}$

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

Vậy $G_{\min}=4$. Giá trị này đạt tại $x-1=\frac{1}{x-1}$

$\Leftrightarrow x=0$ hoặc $x=2$

2.

Áp dụng BĐT Cô-si:

$H=x+\frac{1}{x}=(\frac{x}{4}+\frac{1}{x})+\frac{3}{4}x$

$\geq 2\sqrt{\frac{x}{4}.\frac{1}{x}}+\frac{3}{4}x$
$=1+\frac{3}{4}x\geq 1+\frac{3}{4}.2=\frac{5}{2}$ (do $x\geq 2$)

Vậy $H_{\min}=\frac{5}{2}$. Giá trị này đạt tại $x=2$
 

3.

Áp dụng BĐT Cô-si:

$K=x^2+\frac{1}{x}=(\frac{x^2}{54}+\frac{1}{2x}+\frac{1}{2x})+\frac{53}{54}x^2$

$\geq 3\sqrt[3]{\frac{x^2}{54}.\frac{1}{2x}.\frac{1}{2x}}+\frac{53}{54}x^2$
$=\frac{1}{2}+\frac{53}{54}x^2\geq \frac{1}{2}+\frac{53}{54}.3^2=\frac{28}{3}$ (do $x\geq 3$)

Vậy $K_{\min}=\frac{28}{3}$ khi $x=3$