Áp dụng bđt: 2xy \(\le\)(x + y)²/2

khi đó, ta có: \(\sqrt{\frac{a+b}{2ab}}\ge\sqrt{\frac{a+b}{\frac{\left(a+b\right)^2}{2}}}=\sqrt{\frac{2}{a+b}}=\frac{1}{\sqrt{\frac{a+b}{2}}}\ge\frac{1}{\frac{\frac{a+b}{2}+1}{2}}=\frac{4}{a+b+2}\)

CMTT: \(\sqrt{\frac{b+c}{2bc}}\ge\frac{4}{b+c+2}\)

\(\sqrt{\frac{c+a}{2ca}}\ge\frac{4}{c+a+2}\)

=>Đặt A = \(\sqrt{\frac{a+b}{2ab}}+\sqrt{\frac{b+c}{2bc}}+\sqrt{\frac{a+c}{2ac}}\ge\frac{4}{a+b+2}+\frac{4}{b+c+2}+\frac{4}{a+c+2}\)

Áp dụng bđt svacso : \(\frac{x_1^2}{y_1}+\frac{x_2^2}{y_2}+\frac{x_3^2}{y_3}\ge\frac{\left(x_1+x_2+x_3\right)^2}{y_1+y_2+y_3}\)

 ta có: 

\(A\ge\frac{\left(2+2+2\right)^2}{a+b+2+b+c+2+a+c+2}=\frac{36}{2\left(a+b+c\right)+6}=\frac{36}{12}=3\)

=> Đpcm

a, \(a+11⋮a+3\)

\(a+3+8⋮a+3\)

\(8⋮a+3\)hay \(a+3\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

b, \(a-3⋮a-14\)

\(a-14+11⋮a-14\)

\(11⋮a-14\)hay \(a-14\inƯ\left(11\right)\left\{\pm1;\pm11\right\}\)

a, \(\frac{x-2}{x+1}=\frac{x-3}{x+2}ĐK:x\ne-1;-2\)

\(\Leftrightarrow x^2-4=\left(x-3\right)\left(x+1\right)\Leftrightarrow x^2-4=x^2+x-3x-3\)

\(\Leftrightarrow2x-1=0\Leftrightarrow x=\frac{1}{2}\)

b, \(\frac{2x+1}{x-3}=\frac{2x-3}{x+1}ĐK:x\ne3;-1\)

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

\(\Leftrightarrow2x^2+2x+x+1=2x^2-6x-3x+9\)

\(\Leftrightarrow2x^2+3x+1-2x^2+9x-9=0\)

\(\Leftrightarrow12x-8=0\Leftrightarrow x=\frac{2}{3}\)

Ta luôn có \(4\left(x^3+y^3\right)\ge\left(x+y\right)^3\)(*)

Thật vậy: (*)\(\Leftrightarrow3\left(x-y\right)^2\left(x+y\right)\ge0\)*Đúng với mọi x, y thực dương*

\(\Rightarrow\sqrt[3]{4\left(x^3+y^3\right)}\ge x+y\)

Tương tự, ta có: \(\sqrt[3]{4\left(y^3+z^3\right)}\ge y+z,\sqrt[3]{4\left(z^3+x^3\right)}\ge z+x\)

Cộng theo vế ba bất đẳng thức trên, ta được: \(\sqrt[3]{4\left(x^3+y^3\right)}+\sqrt[3]{4\left(y^3+z^3\right)}+\sqrt[3]{4\left(z^3+x^3\right)}\ge2\left(x+y+z\right)\)

Ta cần chứng minh \(\left(x+y+z\right)+\left(\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}\right)\ge6\)

Thật vậy, ta có: \(\left(x+y+z\right)+\left(\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}\right)\ge3\sqrt[3]{xyz}+\frac{3}{\sqrt[3]{xyz}}\ge3.2=6\)

Vậy bất đẳng thức được chứng minh

Đẳng thức xảy ra khi x = y = z 

ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne4\\x\ne9\end{cases}}\)

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

\(=\left(\frac{x-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\frac{x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\right)\times\frac{2\left(\sqrt{x}-2\right)}{1}\)

\(=\left(\frac{x-4-x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\right)\times\frac{2\left(\sqrt{x}-2\right)}{1}\)

\(=\frac{5}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\times\frac{2\left(\sqrt{x}-2\right)}{1}\)

\(=\frac{5\times2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\frac{10}{\sqrt{x}-3}\)

\(\left(\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{\sqrt{x}+3}{\sqrt{x}-2}\right):\frac{1}{2\sqrt{x}-4}\)

\(=\left(\frac{x-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}-\frac{x-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\right):\frac{1}{2\sqrt{x}-4}\)

\(=\frac{5}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}.\frac{2\left(\sqrt{x}-2\right)}{1}\)

\(=\frac{10\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\frac{10}{\sqrt{x}-3}\)

BĐT cần chứng minh tương đương với: 

\(\frac{a}{b}-\frac{a}{b+c}+\frac{b}{c}-\frac{b}{c+a}+\frac{c}{a}-\frac{c}{a+b}\ge\frac{3}{2}\)

\(\Leftrightarrow\frac{ac}{b\left(b+c\right)}+\frac{ba}{c\left(c+a\right)}+\frac{cb}{a\left(a+b\right)}\ge\frac{3}{2}\)

Ta có: 

\(\frac{ac}{b\left(b+c\right)}+\frac{ba}{c\left(c+a\right)}+\frac{cb}{a\left(a+b\right)}\)

\(=\frac{a^2c^2}{abc\left(b+c\right)}+\frac{b^2a^2}{abc\left(c+a\right)}+\frac{c^2b^2}{abc\left(a+b\right)}\)

\(\ge\frac{\left(ab+bc+ca\right)^2}{abc\left(a+b\right)+abc\left(b+c\right)+abc\left(c+a\right)}\)

\(=\frac{\left(ab+bc+ca\right)^2}{2abc\left(a+b+c\right)}\)

Bất đẳng thức cần chứng minh sẽ đúng nếu ta chứng minh được \(\frac{\left(ab+bc+ca\right)^2}{abc\left(a+b+c\right)}\ge3\)

\(\Leftrightarrow\left(ab+bc+ca\right)^2\ge3abc\left(a+b+c\right)\)

Đặt \(ab=x,bc=y,ca=z\)suy ra ta cần chứng minh 

\(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)

\(\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2zx\ge3xy+3yz+3zx\)

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

Vậy bất đẳng thức ban đầu là đúng. 

Dấu \(=\)xảy ra khi \(a=b=c\).

Xét tam giác AMC và tam giác ABM ta có : 

AM chung 

AC = AB 

BM = MC ( vì M là trung điểm )

^AMC = ^AMB ( 2 góc tương ứng )

Vì ^AMB = ^AMC (cmt)

Mà ^AMB + ^AMC = 180^0 ( 2 góc kề bù )

=)) ^AMB = ^AMC = 90^0 

Vậy AM \(\perp\)BC (đpcm)

Xét ΔΔAMB và ΔΔAMC có:

AM chung

AB = AC (gt)

MB = MC (suy từ gt)

=> ΔΔAMB = ΔΔAMC (c.c.c)

\(\Rightarrow\widehat{AMB}=\widehat{AMC}\) ( hai góc tương ứng )

  mà \(\widehat{AMB}+\widehat{AMC}=180^o\) ( kề bù )

\(\Rightarrow\widehat{AMB}=\widehat{AMC}=90^o\)

Do đó AM ⊥ BC.

\(\frac{x+y+z}{2}=\frac{x}{y+z-5}=\frac{y}{x+z+3}=\frac{z}{x+y+2}=\frac{x+y+z}{2\left(x+y+z\right)}=\frac{1}{2}\)

\(\Rightarrow\hept{\begin{cases}x+y+z=1\\2x=y+z-5\\2y=x+z+3\end{cases}\Leftrightarrow\hept{\begin{cases}x=-\frac{4}{3}\\y=\frac{4}{3}\\z=1\end{cases}}}\)