Lời giải:

Xét tử :

\(\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}=\frac{a^2}{a^2+bc+(-ab-ac)}+\frac{b^2}{b^2+ac+(-ab-bc)}+\frac{c^2}{c^2+ab+(-bc-ac)}\)

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

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

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

\(=\frac{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}=1\)

Xét mẫu (tương tự bên tử)

\(\frac{bc}{a^2+2bc}+\frac{ac}{b^2+2ac}+\frac{ab}{c^2+2ab}=\frac{bc}{(a-c)(a-b)}+\frac{ac}{(b-a)(b-c)}+\frac{ab}{(c-a)(c-b)}\)

\(=\frac{bc(c-b)+ac(a-c)+ab(b-a)}{(a-b)(b-c)(c-a)}=\frac{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}{(a-b)(b-c)(c-a)}\)

\(=\frac{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}{(ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)}=1\)

Do đó:

\(A=\frac{1}{1}=1\)

\(\Leftrightarrow\dfrac{2+3\left(2a+b+2\sqrt{2bc}\right)}{2a+b+2\sqrt{2bc}}\ge\dfrac{16}{\sqrt{2b^2+2\left(a+c\right)^2}+3}\)

\(\Leftrightarrow3+\dfrac{2}{2a+b+2\sqrt{2bc}}\ge\dfrac{16}{\sqrt{2b^2+2\left(a+c\right)^2}+3}\)

Do \(\dfrac{2}{2a+b+2\sqrt{2bc}}\ge\dfrac{2}{2a+b+b+2c}=\dfrac{1}{a+b+c}\)

Và \(2b^2+2\left(a+c\right)^2\ge\left(a+b+c\right)^2\)

Nên ta chỉ cần chứng minh:

\(3+\dfrac{1}{a+b+c}\ge\dfrac{16}{a+b+c+3}\)

Thật vậy, ta có:

\(3+\dfrac{1}{a+b+c}=\dfrac{1}{1}+\dfrac{1}{1}+\dfrac{1}{1}+\dfrac{1}{a+b+c}\ge\dfrac{16}{1+1+1+a+b+c}=\dfrac{16}{a+b+c+3}\) (đpcm)

Dấu "=" xảy ra khi \(a=\dfrac{b}{2}=c=\dfrac{1}{4}\)

cho mình hỏi bạn biết làm chưa nếu rồi thì giúp mình được không ạ mình ko biết làm

Câu a :

Theo BĐT cauchy schwar ta có :

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(1+1+1\right)^2}{x+y+z}=\dfrac{9}{x+y+z}\)

\(\Rightarrow\left(x+y+z\right)\left(\dfrac{9}{x+y+z}\right)\ge9\)

Câu b : Sửa lại đề nha :

Theo BĐT cauchy schwar ta có :

\(\dfrac{1}{a^2+2bc}+\dfrac{1}{b^2+2ca}+\dfrac{1}{c^2+2ab}\ge\dfrac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2\left(ab+bc+ca\right)}=\dfrac{9}{\left(a+b+c\right)^2}\)

Vì \(a+b+c\le\Rightarrow\left(a+b+c\right)^2\le1\)

\(\Rightarrow\) \(\dfrac{9}{\left(a+b+c\right)^2}\ge9\)

Bài 2 :

Ta có : \(4p(p-a)\)\(=2\left(a+b+c\right)\left(\dfrac{a+b+c}{2}-a\right)\)

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

\(=\left(a+b+c\right)\left(b+c-a\right)\)

\(=ab+ac-a^2+b^2+bc-ab+bc+c^2-ac\)

\(=2bc+b^2+c^2-a^2\left(dpcm\right)\)

Vậy :

Bai 2:

Ta có:

\(VP=4p\left(p-a\right)=2p.2p-2a.2p\) (1)

Thay \(a+b+c=2p\) vào (1) ta có:

\(\left(a+b+c\right)^2-2a.\left(a+b+c\right)\)

\(=a^2+b^2+c^2+2ab+2ac+2bc-2a^2-2ab-2ac\)

\(=-a^2+b^2+c^2+2bc=VT\)

Vậy \(2bc+b^2+c^2-a^2=4p\left(p-a\right)\)

Chúc bạn học tốt!!!

Giải giùm bài 1 đi cháu Trần Nguyễn Bảo Quyên

Xét biểu thức \(x+y+xy+1=\left(x+1\right)\left(y+1\right)\)

Từ giả thiết suy ra \(x+1=\dfrac{\left(b+c\right)^2-a^2}{2bc};y+1=\dfrac{4bc}{\left(b+c\right)^2-a^2}\)

Do đó \(\left(x+1\right)\left(y+1\right)=2\Rightarrow xy+x+y+1=2\Rightarrow xy+x+y=1\)

A = x + y + xy

A = x( y + 1) + y

A = \(\dfrac{b^2+c^2-a^2}{2bc}\left(\dfrac{a^2-b^2+2bc-c^2}{\left(b+c\right)^2-a^2}+1\right)+\dfrac{a^2-\left(b-c\right)^2}{\left(b+c\right)^2-a^2}\)

A = \(\dfrac{b^2+c^2-a^2}{2bc}.\dfrac{4bc}{\left(b+c\right)^2-a^2}+\dfrac{a^2-\left(b-c\right)^2}{\left(b+c\right)^2-a^2}\)

A= \(\dfrac{2\left(b^2+c^2-a^2\right)+a^2-b^2+2bc-c^2}{\left(b+c\right)^2-a^2}\)

A = \(\dfrac{b^2+2bc+c^2-a^2}{\left(b+c\right)^2-a^2}=\dfrac{\left(b+c\right)^2-a^2}{\left(b+c\right)^2-a^2}=1\)