§1. Bất đẳng thức

** Bài này chỉ đúng khi $a,b,c$ không âm thôi bạn nhé.

Lời giải:
Theo BĐT Schur:

$a^3+b^3+c^3+3abc\geq ab(a+b)+bc(b+c)+ca(c+a)$

$\Rightarrow a^3+b^3+c^3+6abc\geq (a+b+c)(ab+bc+ac)$

$\Leftrightarrow a^3+b^3+c^3+3(a+b+c)(ab+bc+ac)+6abc\geq 4(a+b+c)(ab+bc+ac)$
$\Leftrightarrow a^3+b^3+c^3+3[(a+b)(b+c)(c+a)+abc]+6abc\geq 4(a+b+c)(ab+bc+ac)$

$\Leftrightarrow a^3+b^3+c^3+3(a+b)(b+c)(c+a)+9abc\geq 4(a+b+c)(ab+bc+ac)$

$\Leftrightarrow (a+b+c)^3+9abc\geq 4(a+b+c)(ab+bc+ac)$

Dấu "=" xảy ra khi $a=b=c$

Em tách nhỏ đăng 1-2 câu/1 lần hỏi thôi nha!

Đặt \(\left(a+1;b+1;c+1\right)=\left(x;y;z\right)\Rightarrow1\le x\le y\le z\le2\)

\(B=\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{z}{x}+\dfrac{x}{z}+3\) (1)

Do \(x\le y\le z\Rightarrow\left(z-y\right)\left(y-x\right)\ge0\)

\(\Leftrightarrow xy+yz\ge y^2+zx\)

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

Tương tự: \(1+\dfrac{z}{x}\ge\dfrac{y}{x}+\dfrac{z}{y}\)

Cộng vế: \(2+\dfrac{x}{z}+\dfrac{z}{x}\ge\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{y}{x}\) (2)

Từ (1); (2) \(\Rightarrow B\le2\left(\dfrac{x}{z}+\dfrac{z}{x}\right)+5\)

Đặt \(\dfrac{z}{x}=t\Rightarrow1\le t\le2\)

\(\Rightarrow B\le2\left(t+\dfrac{1}{t}\right)+5=\dfrac{2t^2+2}{t}+5=\dfrac{2t^2+2}{t}-5+10\)

\(\Rightarrow B\le\dfrac{2t^2-5t+2}{t}+10=\dfrac{\left(t-2\right)\left(2t-1\right)}{t}+10\le10\)

\(B_{max}=10\) khi \(t=2\) hay \(\left(a;b;c\right)=\left(0;0;1\right);\left(0;1;1\right)\)

Áp dụng bất đẳng thức Cô-si với số x>0

Ta có :

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

Vậy min của A là 2 khi \(x = \dfrac{1}{x} \Leftrightarrow x = 1\)

\(\dfrac{x+1}{x}\) hay \(x+\dfrac{1}{x}\) ạ ?

\(\sqrt{\dfrac{a}{b+c}}=\dfrac{a}{\sqrt{a\left(b+c\right)}}\ge\dfrac{2a}{a+b+c}\)

Tương tự: \(\sqrt{\dfrac{b}{c+a}}\ge\dfrac{2b}{a+b+c}\) ; \(\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2c}{a+b+c}\)

Cộng vế:

\(VT\ge\dfrac{2a+2b+2c}{a+b+c}=2\)

Dấu "=" ko xảy ra nên \(VT>2\)

Áp dụng cosi:

`x^2+y^2>=2xy`

`=>x^2+y^2>=2.7=14`

`=>` Chọn C.14

\(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}\ge\dfrac{3}{\sqrt[3]{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)

\(\dfrac{a}{1+a}+\dfrac{b}{1+b}+\dfrac{c}{1+c}\ge\dfrac{3\sqrt[3]{abc}}{\sqrt[3]{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)

Cộng vế với vế:

\(3\ge\dfrac{3+3\sqrt[3]{abc}}{\sqrt[3]{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)

\(\Leftrightarrow\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge\left(1+\sqrt[3]{abc}\right)^3\)

\(\sqrt{a}+\sqrt{a}+\sqrt[3]{b}+\sqrt[3]{b}+\sqrt[3]{b}+\sqrt[4]{c}+\sqrt[4]{c}+\sqrt[4]{c}+\sqrt[4]{c}\ge9\sqrt[9]{\left(\sqrt{a}\right)^2\left(\sqrt[3]{b}\right)^3\left(\sqrt[4]{c}\right)^4}\)

\(\Leftrightarrow2\sqrt{a}+3\sqrt[3]{b}+4\sqrt[4]{c}\ge9\sqrt[9]{abc}\)

Hình như thế này mới đúng chứ \(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\)

Áp dụng BĐT Cosi:

\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2.\dfrac{a}{c};\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2.\dfrac{b}{a};\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2.\dfrac{c}{b}\)

\(\Rightarrow2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}\right)\)

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

Đẳng thức xảy ra khi \(a=b=c>0\)