a) Bất đẳng thức đúng khi a = b = 2c

do đó \(\sqrt{c\left(2c-c\right)}+\sqrt{c\left(2c-c\right)}\le n\sqrt{2c.2c}\Leftrightarrow n\ge1\)

xảy ra khi n = 1

Thật vậy, ta có :

\(\sqrt{\frac{c}{b}.\frac{a-c}{a}}+\sqrt{\frac{c}{a}.\frac{b-c}{b}}\le\frac{1}{2}\left(\frac{c}{b}+\frac{a-c}{a}+\frac{c}{a}+\frac{b-c}{b}\right)\)

\(\Leftrightarrow\sqrt{c\left(a-c\right)}+\sqrt{c\left(b-c\right)}\le\sqrt{ab}\)

Vậy n nhỏ nhất là 1

b) Ta có : a + b = \(\sqrt{\left(a+b\right)^2}\le\sqrt{\left(a+b\right)^2+\left(a-b\right)^2}=\sqrt{2\left(a^2+b^2\right)}\)

Áp dụng, ta được : \(\sqrt{1}+\sqrt{n}\le\sqrt{2\left(n+1\right)},\sqrt{2}+\sqrt{n-1}\le\sqrt{2\left(1+n\right)},...\)

\(\sqrt{n}+\sqrt{1}\le\sqrt{2\left(1+n\right)};\sqrt{n-1}+\sqrt{2}\le\sqrt{2\left(1+n\right)},...\)

\(\sqrt{1}+\sqrt{n}\le\sqrt{2\left(1+n\right)}\)

do đó : \(4\left(\sqrt{1}+\sqrt{2}+...+\sqrt{n}\right)\le2n\sqrt{2\left(1+n\right)}\)

\(\Rightarrow\sqrt{1}+\sqrt{2}+...+\sqrt{n}\le n\sqrt{\frac{n+1}{2}}\)

2. \(BĐT\Leftrightarrow\frac{1}{1+\frac{2}{a}}+\frac{1}{1+\frac{2}{b}}+\frac{1}{1+\frac{2}{c}}\ge1\)

Đặt\(\frac{2}{a}=x;\frac{2}{b}=y;\frac{2}{c}=z\)thì \(\hept{\begin{cases}x,y,z>0\\xyz=8\end{cases}}\)

Ta cần chứng minh \(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\ge1\Leftrightarrow\left(yz+y+z+1\right)+\left(zx+z+x+1\right)+\left(xy+x+y+1\right)\ge xyz+\left(xy+yz+zx\right)+\left(x+y+z\right)+1\)\(\Leftrightarrow x+y+z\ge6\)(Đúng vì \(x+y+z\ge3\sqrt[3]{xyz}=6\))

Đẳng thức xảy ra khi x = y = z = 2 hay a = b = c = 1

3. Ta có: \(a+b+c\le\sqrt{3}\Rightarrow\left(a+b+c\right)^2\le3\)

Ta có đánh giá quen thuộc \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

Từ đó suy ra \(ab+bc+ca\le1\)

\(A=\frac{\sqrt{a^2+1}}{b+c}+\frac{\sqrt{b^2+1}}{c+a}+\frac{\sqrt{c^2+1}}{a+b}\ge\frac{\sqrt{a^2+ab+bc+ca}}{b+c}+\frac{\sqrt{b^2+ab+bc+ca}}{c+a}+\frac{\sqrt{c^2+ab+bc+ca}}{a+b}\)\(=\frac{\sqrt{\left(a+b\right)\left(a+c\right)}}{b+c}+\frac{\sqrt{\left(b+a\right)\left(b+c\right)}}{c+a}+\frac{\sqrt{\left(c+a\right)\left(c+b\right)}}{a+b}\ge3\sqrt[3]{\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=3\)Đẳng thức xảy ra khi \(a=b=c=\frac{1}{\sqrt{3}}\)

1. Ta thấy:

\(\frac{(a-b)^3}{(\sqrt{a}-\sqrt{b})^3}-b\sqrt{b}+2a\sqrt{a}=\frac{(\sqrt{a}-\sqrt{b})^3(\sqrt{a}+\sqrt{b})^3}{(\sqrt{a}-\sqrt{b})^3}-b\sqrt{b}+2a\sqrt{a}\)

\(=(\sqrt{a}+\sqrt{b})^3-b\sqrt{b}+2a\sqrt{a}=a\sqrt{a}+b\sqrt{b}+3\sqrt{ab}(\sqrt{a}+\sqrt{b})-b\sqrt{b}+2a\sqrt{a}\)

\(=3a\sqrt{a}+3\sqrt{ab}(\sqrt{a}+\sqrt{b})=3\sqrt{a}(a+\sqrt{ab}+b)\)

$a\sqrt{a}-b\sqrt{b}=(\sqrt{a}-\sqrt{b})(a+\sqrt{ab}+b)$

\(\frac{\frac{(a-b)^3}{(\sqrt{a}-\sqrt{b})^3}-b\sqrt{b}+2a\sqrt{a}}{a\sqrt{a}-b\sqrt{b}}=\frac{3\sqrt{a}}{\sqrt{a}-\sqrt{b}}(1)\)

\(\frac{3a+3\sqrt{ab}}{b-a}=\frac{3\sqrt{a}(\sqrt{a}+\sqrt{b})}{(\sqrt{b}-\sqrt{a})(\sqrt{b}+\sqrt{a})}=\frac{-3\sqrt{a}}{\sqrt{a}-\sqrt{b}}(2)\)

Từ $(1);(2)$ ta có đpcm.

Câu 2:

Điều kiện đã cho tương đương với:

$\frac{a-b}{a(a+b)}+\frac{a+b}{a(a-b)}=\frac{3a-b}{(a-b)(a+b)}$

$\Leftrightarrow \frac{(a-b)^2}{a(a+b)(a-b)}+\frac{(a+b)^2}{a(a-b)(a+b)}=\frac{a(3a-b)}{a(a-b)(a+b)}$

$\Leftrightarrow (a-b)^2+(a+b)^2=a(3a-b)$

$\Leftrightarrow 2a^2+2b^2=3a^2-ab$

$\Leftrightarrow a^2-ab-2b^2=0$

$\Leftrightarrow (a+b)(a-2b)=0$

$\Leftrightarrow a=-b$ hoặc $a=2b$

Nếu $a=-b$ thì $|a|=|b|$ (trái giả thiết). Do đó $a=2b$

Khi đó:

$P=\frac{(2b)^3+2(2b)^2.b+3b^3}{2(2b)^3+2b.b^2+b^3}=\frac{19b^3}{19b^3}=1$

1. 

\(DK:x\ge2\)

\(\Leftrightarrow\left(3\sqrt{x-2}-3\right)+\left(3-\sqrt{x+6}\right)-\left(2x-6\right)=0\)

\(\Leftrightarrow\frac{3\left(x-3\right)}{\sqrt{x-2}+3}-\frac{x-3}{3+\sqrt{x+6}}-2\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{3}{\sqrt{x-2}+3}-\frac{1}{3+\sqrt{x+6}}-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\left(1\right)\\\frac{3}{\sqrt{x-2}+3}-\frac{1}{3+\sqrt{x+6}}-2=0\left(2\right)\end{cases}}\)

PT(2) khac khong voi moi \(x\ge2\)

Vay nghiem cua PT la \(x=3\)

\(x^3+2x=y^2-2009\)

\(\Leftrightarrow x^3-x=y^2-3x-2009\)

\(\Leftrightarrow\left(x-1\right)x\left(x+1\right)=y^2-3x-2009\)

Dễ thấy VT chia hết cho 3 nên VP chia hết cho 3 

Suy ra \(y^2\) chia 3 dư 2 vì 2009 chia 3 dư 2 và 3x chia hết cho 3 ( vô lý vì số chính phương ko chia 3 dư 2 ) 

Vậy pt vô nghiệm

\(\frac{1}{a}+\frac{1}{b}-\left(\sqrt{\frac{a}{b}}-\sqrt{\frac{b}{a}}\right)^2\ge2\sqrt{2}\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}-\frac{a}{b}-\frac{b}{a}+2\ge2\sqrt{2}\)

\(\Leftrightarrow\frac{a+b}{ab}-\frac{a^2+b^2}{ab}+2-2\sqrt{2}\ge0\)

\(\Leftrightarrow\frac{a+b-1}{ab}+2-2\sqrt{2}\ge0\left(1\right)\)

Áp dụng BĐT phụ \(ab\le\frac{a^2+b^2}{2}=\frac{1}{2}\)

\(\left(1\right)\Leftrightarrow2a+2b-2+2-2\sqrt{2}\ge0\) 

Áp dụng 1 lần nữa ta có:

\(\left(1\right)\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\ge0\left(true\right)\)

P/S:E ko chắc ạ

1,

\(\frac{a}{1+\frac{b}{a}}+\frac{b}{1+\frac{c}{b}}+\frac{c}{1+\frac{a}{c}}=\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{a+b+c}{2}\ge\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2}=\frac{2}{2}=1\left(Q.E.D\right)\)