Áp dụng bất đẳng thức AM-GM ta có :

\(B=\frac{12}{x-1}+\frac{x-1+1}{3}=\frac{12}{x-1}+\frac{x-1}{3}+\frac{1}{3}\ge2\sqrt{\frac{12}{x-1}\cdot\frac{x-1}{3}}+\frac{1}{3}=4+\frac{1}{3}=\frac{13}{3}\)

Dấu "=" xảy ra <=> \(\frac{12}{x-1}=\frac{x-1}{3}\Rightarrow x=7\left(x\ge1\right)\). Vậy MinB = 13/3

a) giả sử \(x\ge y\ge3\)

P(x)=x+1/x

P(y)=y+1/y

P(x)-p(y)=(x+1/x)-(y+1/y)=(x-y)+(1/x-1/y)=A

\(x\ge y\ge3\Rightarrow\frac{1}{x}\le\frac{1}{y}\hept{\begin{cases}x-y\le0\\\frac{1}{x}-\frac{1}{y}\le0\end{cases}\Rightarrow A\le0}\)

Kết luận a cành lớn thì P(a) càng lớn

=> P_min=P(3)=3+1/3=10/3

Ok ta cần chứng minh A>=0

\(A=\left(x-y\right)+\left(\frac{1}{x}-\frac{1}{y}\right)=\left(x-y\right)+\frac{\left(y-x\right)}{xy}=\left(x-y\right)-\frac{\left(x-y\right)}{xy}\\ \)

\(A=\left(x-y\right)\left[1-\frac{1}{xy}\right]\)

\(x\ge y\ge3\Rightarrow\hept{\begin{cases}x-y\ge0\\xy\ge9\\\frac{1}{xy}\le\frac{1}{9}< 1\Rightarrow1-\frac{1}{xy}>0\end{cases}}\Rightarrow A\ge0\)

a.

Xét x₁ > x₂ >= 3

P(x₁) - P(x₂) = x₁ - x₂ + \(\frac{1}{x_1}\) - \(\frac{1}{x_2}\)

= \(\frac{x_1^2x_2-x_1x_2^2+x_2-x_1}{x_1x_2}\)

= \(\frac{x_1x_2\left(x_1-x_2\right)+x_2-x_1}{x_1x_2}\)

= \(\frac{\left(x_1-x_2\right)\left(x_1x_2-1\right)}{x_1x_2}\)> 0 (vì x₁ > x₂. x₁x₂ > 9)

Vậy, với a > 3 => a+1/a > 3+1/3

=> Min_P = 3+1/3

b. Xét x > y >= 2

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

= \(\frac{x^3y^2-x^2y^3+y^2-x^2}{x^2y^2}\)

= \(\frac{x^2y^2\left(x-y\right)+\left(y-x\right)\left(y+x\right)}{x^2y^2}\)

= \(\frac{\left(x-y\right)\left(x^2y^2-\left(x+y\right)\right)}{x^2y^2}\)> 0

Vì x > y >= 2

đặt x = 2+a (a>0)

y = 2+b (b>=0)

=> (2+a)²(2+b)² - 4 - a - b > 0 (Lấy bình phương rồi nhân vào sẽ rút được dấu - tại - 4 - a - b)

Vậy, với x > y >= 2 => S(x) > S(y)

=> Min_S = S(2) = 2 + 1/4.

Xét \(\dfrac{a}{a^2+1}+\dfrac{3\left(a-2\right)}{25}-\dfrac{2}{5}=\dfrac{a}{a^2+1}+\dfrac{3a-16}{25}=\dfrac{\left(3a-4\right)\left(a-2\right)^2}{25\left(a^2+1\right)}\ge0\)

\(\Rightarrow\dfrac{a}{a^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(a-2\right)}{25}\)

CMTT \(\Rightarrow\left\{{}\begin{matrix}\dfrac{b}{b^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(b-2\right)}{25}\\\dfrac{c}{c^2+1}\ge\dfrac{2}{5}-\dfrac{3\left(c-2\right)}{25}\end{matrix}\right.\)

Cộng vế theo vế:

\(\Rightarrow VT\ge\dfrac{2}{5}+\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{3\left(a-2\right)+3\left(b-2\right)+3\left(c-2\right)}{25}\ge\dfrac{6}{5}-\dfrac{3\left(a+b+c-6\right)}{25}=\dfrac{6}{5}\)

Dấu \("="\Leftrightarrow a=b=c=2\)

\(P=\dfrac{5x}{2}+\dfrac{x}{2}+\dfrac{1}{2x}\ge\dfrac{5x}{2}+2\sqrt{\dfrac{x}{2}.\dfrac{1}{2x}}\ge\dfrac{5.1}{2}+2.\dfrac{1}{2}=\dfrac{7}{2}\)

\(\Rightarrow P_{min}=\dfrac{7}{2}\) khi \(x=1\)

Ta có: 

Vì \(\frac{2}{3}< x< \frac{13}{2}\Rightarrow\hept{\begin{cases}3x-2>0\\10-x>0\\13-2x>0\end{cases}}\)

Khi đó: \(\frac{1}{3x-2}-\frac{1}{x-10}+\frac{1}{13-2x}\)

\(=\frac{1}{3x-2}+\frac{1}{10-x}+\frac{1}{13-2x}\) \(\left(1\right)\)

Áp dụng BĐT Cauchy Schwarz ta được:

\(\left(1\right)\ge\frac{\left(1+1+1\right)^2}{3x-2+10-x+13-2x}\)

\(=\frac{3^2}{21}=\frac{3}{7}\)

Vậy với \(\frac{2}{3}< x< \frac{13}{2}\) thì \(\frac{1}{3x-2}-\frac{1}{x-10}+\frac{1}{13-2x}\ge\frac{3}{7}\)