1)Từ đề bài:

`=>a^2+4b+4+b^2+4c+4+c^2+4a+4=0`

`<=>(a+2)^2+(b+2)^2+(c+2)^2=0`

`<=>a=b=c-2`

`ab+bc+ca=abc`

`<=>1/a+1/b+1/c=1`

`<=>(1/a+1/b+1/c)^2=1`

`<=>1/a^2+1/b^2+1/c^2+2/(ab)+2/(bc)+2/(ca)=1`

`<=>1/a^2+1/b^2+1/c^2=1-(2/(ab)+2/(bc)+2/(ca))`

`a+b+c=0`

Chia 2 vế cho `abc`

`=>1/(ab)+1/(bc)+1/(ca)=0`

`=>2/(ab)+2/(bc)+2/(ca)=0`

`=>1/a^2+1/b^2+1/c^2=1-0=1`

Từng sau em hạn chế đăng nhiều bài cùng một lúc như thế này nhé. 

Bài 1:

Ta có: \(a+\frac{4}{(a-b)(b+1)^2}=(a-b)+\frac{b+1}{2}+\frac{b+1}{2}+\frac{4}{(a-b)(b+1)^2}-1\)

Áp dụng BĐT AM-GM cho các số không âm ta có:

\((a-b)+\frac{b+1}{2}+\frac{b+1}{2}+\frac{4}{(a-b)(b+1)^2}\geq 4\sqrt[4]{\frac{4(a-b)(b+1)^2}{4(a-b)(b+1)^2}}=4\)

\(\Rightarrow a+\frac{4}{(a-b)(b+1)^2}=(a-b)+\frac{b+1}{2}+\frac{b+1}{2}+\frac{4}{(a-b)(b+1)^2}-1\geq 4-1\)

\(\Leftrightarrow a+\frac{4}{(a-b)(b+1)^2}\geq 3\)

Ta có đpcm.

Dấu bằng xảy ra khi \(a-b=\frac{b+1}{2}=\frac{4}{(a-b)(b+1)^2}\)

\(\Leftrightarrow a=2; b=1\)

Bài 2:

Đặt \(\left(\frac{a}{b}, \frac{b}{c}, \frac{c}{a}\right)\mapsto (x,y,z)\Rightarrow xyz=1\)

BĐT cần chứng minh tương đương với:

\(x^2+y^2+z^2\geq \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)

\(\Leftrightarrow x^2+y^2+z^2\geq \frac{xy+yz+xz}{xyz}=xy+yz+xz(*)\)

Áp dụng BĐT AM-GM:

\(x^2+y^2\geq 2\sqrt{x^2y^2}=2xy\)

\(y^2+z^2\geq 2\sqrt{y^2z^2}=2yz\)

\(z^2+x^2\geq 2\sqrt{z^2x^2}=2zx\)

Cộng theo vế: \(\Rightarrow 2(x^2+y^2+z^2)\geq 2(xy+yz+xz)\)

\(\Leftrightarrow x^2+y^2+z^2\geq xy+yz+xz\)

Do đó (*) đúng, ta có đpcm.

Dấu bằng xảy ra khi \(x=y=z=1\Leftrightarrow a=b=c\)

Bài 3:

Ta có: \(\text{VT}=(\frac{b}{\sqrt{a}}+\frac{c}{\sqrt{b}}+\frac{a}{\sqrt{c}})+(\frac{c}{\sqrt{a}}+\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}})\)

Áp dụng BĐT Bunhiacopxky:

\((\frac{b}{\sqrt{a}}+\frac{c}{\sqrt{b}}+\frac{a}{\sqrt{c}})(\sqrt{a}+\sqrt{b}+\sqrt{c})\geq (\sqrt{b}+\sqrt{c}+\sqrt{a})^2\)

\(\frac{b}{\sqrt{a}}+\frac{c}{\sqrt{b}}+\frac{a}{\sqrt{c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}(1)\)

Áp dụng BĐT AM-GM:

\(\frac{c}{\sqrt{a}}+\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{c}}\geq 3\sqrt[3]{\frac{abc}{\sqrt{abc}}}=3(2)\) do $abc=1$

Từ \((1); (2)\Rightarrow \text{VT}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3\) (đpcm)

Dấu bằng xảy ra khi \(a=b=c=1\)

Bài 4:

Vì $a,b,c$ là ba cạnh tam giác nên \(b+c-a, c+a-b, a+b-c>0\)

Áp dụng BĐT AM-GM ta có:

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

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

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

Cộng theo vế và rút gọn:

\(\frac{a^2}{b+c-a}+\frac{b^2}{c+a-b}+\frac{c^2}{a+b-c}+a+b+c\geq 2(a+b+c)\)

\(\Leftrightarrow \frac{a^2}{b+c-a}+\frac{b^2}{c+a-b}+\frac{c^2}{a+b-c}\geq a+b+c\)

Ta có đpcm.

Dấu bằng xảy ra khi \(a=b=c\)

2: Ta có: \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=\dfrac{a\left(a+b+c\right)}{b+c}+\dfrac{b\left(a+b+c\right)}{c+a}+\dfrac{c\left(a+b+c\right)}{a+b}-a-b-c=\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=a+b+c-a-b-c=0\)

1: Sửa đề: Cho \(x,y,z\ne0\) và \(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}=\dfrac{2}{2x+y+2z}\).

CM:....

Đặt 2x = x', 2z = z'.

Ta có: \(\dfrac{2}{x'}+\dfrac{2}{y}+\dfrac{2}{z'}=\dfrac{2}{x'+y+z'}\)

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

\(\Leftrightarrow\dfrac{1}{x'}-\dfrac{1}{x'+y+z'}+\dfrac{1}{y}+\dfrac{1}{z'}=0\)

\(\Leftrightarrow\dfrac{y+z'}{x'\left(x'+y+z'\right)}+\dfrac{y+z'}{yz'}=0\)

\(\Leftrightarrow\dfrac{\left(y+z'\right)\left(yz'+x'^2+x'y+x'z'\right)}{x'yz'\left(x'+y+z'\right)}=0\)

\(\Leftrightarrow\dfrac{\left(x'+y\right)\left(y+z'\right)\left(z'+x'\right)}{x'yz'\left(x'+y+z'\right)}=0\Leftrightarrow\left(2x+y\right)\left(y+2z\right)\left(2z+2x\right)=0\Leftrightarrow\left(2x+y\right)\left(y+2z\right)\left(z+x\right)=0\left(đpcm\right)\)

\(1,a+b+c=0\Leftrightarrow a=-b-c\Leftrightarrow a^2=b^2+2bc+c^2\Leftrightarrow b^2+c^2=a^2-2bc\)

Tương tự: \(\left\{{}\begin{matrix}a^2+b^2=c^2-2ab\\c^2+a^2=b^2-2ac\end{matrix}\right.\)

\(\Leftrightarrow N=\dfrac{a^2}{a^2-a^2+2bc}+\dfrac{b^2}{b^2-b^2+2ca}+\dfrac{c^2}{c^2-c^2+2ac}\\ \Leftrightarrow N=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2bc}=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{a^3+b^3+c^3-3abc+3abc}{2abc}\\ \Leftrightarrow N=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc}{2abc}\\ \Leftrightarrow N=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)