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 \("="\Leftrightarrow a=b=c=2\)

Lời giải:

Xét:

$\frac{a}{a^2+1}-\left(\frac{16}{25}-\frac{3}{25}a\right)=\frac{(a-2)^2(3a-4)}{25(a^2+1)}\geq 0$ với mọi $a\geq \frac{4}{3}$

$\Rightarrow \frac{a}{a^2+1}\geq \frac{16}{25}-\frac{3}{25}a$

Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế, suy ra:

$A\geq \frac{48}{25}-\frac{3}{25}(a+b+c)=\frac{6}{5}$

Vậy $A_{\min}=\frac{6}{5}$.

Giá trị này đạt tại $a=b=c=2$

Ta có \(a^2+\dfrac{1}{b+c}=a^2+\dfrac{1}{6-a}\)

Mà \(a+b+c=6\Rightarrow0\le a,b,c\le2\)

\(\Rightarrow a^2+\dfrac{1}{6-a}\ge2^2+\dfrac{1}{6-2}=\dfrac{17}{4}\)

\(\Rightarrow P=\sum\sqrt{a^2+\dfrac{1}{b+c}}=\sum\sqrt{a^2+\dfrac{1}{6-a}}\ge\sqrt{\dfrac{17}{4}}+\sqrt{\dfrac{17}{4}}+\sqrt{\dfrac{17}{4}}=\dfrac{3\sqrt{17}}{2}\)

Dấu \("="\Leftrightarrow a=b=c=2\)

1. Vì x, y, z > 0

\(xy+yz+zx\ge2xyz\)

\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge2\)

Suy ra:

\(\dfrac{1}{x}\ge1-\dfrac{1}{y}+1-\dfrac{1}{z}=\dfrac{y-1}{y}+\dfrac{z-1}{z}\ge2\sqrt{\dfrac{\left(y-1\right)\left(z-1\right)}{yz}}\). (1)

Tương tự \(\dfrac{1}{y}\ge2\sqrt{\dfrac{\left(z-1\right)\left(x-1\right)}{zx}}\) (2)

và \(\dfrac{1}{z}\ge2\sqrt{\dfrac{\left(x-1\right)\left(y-1\right)}{xy}}\) (3)

Nhân (1), (2), (3) với nhau theo vế ta được

\(\dfrac{1}{xyz}\ge\dfrac{8\left(x-1\right)\left(y-1\right)\left(z-1\right)}{xyz}\)

\(\Leftrightarrow\left(x-1\right)\left(y-1\right)\left(z-1\right)\le\dfrac{1}{8}\)

Đẳng thức xảy ra \(\Leftrightarrow x=y=z=\dfrac{3}{2}\)

\(\dfrac{c+1}{c+3}\ge\dfrac{1}{a+2}+\dfrac{3}{b+4}\)

\(\Leftrightarrow1-\dfrac{2}{c+3}\ge\dfrac{1}{a+2}+\dfrac{3}{b+4}\)

\(\Leftrightarrow1-\dfrac{1}{a+2}\ge\dfrac{3}{b+4}+\dfrac{2}{c+3}\ge2\sqrt{\dfrac{6}{\left(b+4\right)\left(c+3\right)}}\)

Hay \(\dfrac{a+1}{a+2}\ge2\sqrt{\dfrac{6}{\left(b+4\right)\left(c+3\right)}}\) (1)

Tương tự \(\dfrac{b+1}{b+4}\ge2\sqrt{\dfrac{2}{\left(c+3\right)\left(a+2\right)}}\) (2)

và \(\dfrac{c+1}{c+3}\ge2\sqrt{\dfrac{3}{\left(a+2\right)\left(b+4\right)}}\) (3)

Nhân (1), (2), (3) vế theo vế

\(\dfrac{\left(a+1\right)\left(b+1\right)\left(c+1\right)}{\left(a+2\right)\left(b+4\right)\left(c+3\right)}\ge8.\dfrac{6}{\left(a+2\right)\left(b+4\right)\left(c+3\right)}\)

\(\Leftrightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge48\)

Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=5\\c=3\end{matrix}\right.\)

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.

2)a)\(2\left(a^2+b^2\right)\ge\left(a+b\right)^2\Rightarrow\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\)

c)\(a^3+b^3-a^2b-ab^2=a^2\left(a-b\right)-b^2\left(a-b\right)=\left(a-b\right)^2\left(a+b\right)\ge0\\ \Leftrightarrow a^3+b^3\ge ab\left(a+b\right)\)

b)\(a^3+b^3\ge a^2b+ab^2\Leftrightarrow4a^3+4b^3\ge a^3+b^3+3a^b+3ab^2\\ \Leftrightarrow4\left(a^3+b^3\right)\ge\left(a+b\right)^3\Leftrightarrow\dfrac{a^3+b^3}{2}\ge\left(\dfrac{a+b}{2}\right)^3\)

\(x+2y=4\Leftrightarrow x=4-2y\)

\(\Rightarrow xy=y\left(4-2y\right)=-2y^2+4y=-2\left(y-1\right)^2+2\le2\)

Vậy max M là 2 khi y=1, x= 2

2)Tương tự

Lời giải:

Áp dụng BĐT AM-GM:

$P=(a+1)+\frac{2}{a+1}+2\geq 2\sqrt{(a+1).\frac{2}{a+1}}+2=2\sqrt{2}+2$

Vậy $P_{\min}=2\sqrt{2}+2$

Giá trị này đạt tại $(a+1)^2=2; a>0\Leftrightarrow a=\sqrt{2}-1$

------------------------

Bổ sung ĐK: $a>1$

$X=\frac{a^2-1+2}{a-1}=a+1+\frac{2}{a-1}$

$=(a-1)+\frac{2}{a-1}+2$

$\geq 2\sqrt{2}+2$ (AM-GM)

Vậy $X_{\min}=2\sqrt{2}+2$
Giá trị đạt tại $(a-1)^2=\sqrt{2}; a>1\Leftrightarrow a=\sqrt{2}+1$

1a)\(a^2+b^2\ge\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{a^2+b^2}{2}\ge\dfrac{1}{4}\)(1)

Lại có:\(\dfrac{a^2+b^2}{2}\ge\dfrac{\left(a+b\right)^2}{4}=\dfrac{1}{4}\)

\(\Rightarrow\left(1\right)\) đúng\(\Rightarrowđpcm\)

1b)\(a^2+b^2+c^2\ge\dfrac{1}{3}\)

\(\Leftrightarrow\dfrac{a^2}{2}+\dfrac{b^2}{2}+\dfrac{c^2}{2}\ge\dfrac{1}{6}\)(2)

Lại có:\(\dfrac{a^2}{2}+\dfrac{b^2}{2}+\dfrac{c^2}{2}\ge\dfrac{\left(a+b+c\right)^2}{6}=\dfrac{1}{6}\)

\(\Rightarrow\left(2\right)\) đúng\(\Rightarrowđpcm\)

2b)Ta có:\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)(bđt phụ)

\(\Leftrightarrow ab+bc+ca\le\dfrac{4^2}{3}=\dfrac{16}{3}\)

\(\Rightarrow MAXA=\dfrac{16}{3}\Leftrightarrow x=y=z=\dfrac{4}{3}\)