Áp dụng bđt Holder, ta có:

\(\left(\sqrt{\frac{ab}{a^2+b^2}}+\sqrt{\frac{bc}{b^2+c^2}}+\sqrt{\frac{ca}{c^2+a^2}}\right).\left(\sqrt{\frac{ab}{a^2+b^2}}+\sqrt{\frac{bc}{b^2+c^2}}+\sqrt{\frac{ca}{c^2+a^2}}\right)\left[a^2b^2\left(a^2+b^2\right)+b^2c^2\left(b^2+c^2\right)+c^2a^2\left(c^2+a^2\right)\right]\ge\left(ab+bc+ca\right)^3=\frac{\left(a^2+b^2+c^2\right)^3}{8}\)

=>\(VT^2\ge\frac{1}{8}.\frac{\left(a^2+b^2+c^2\right)^3}{a^2b^4+a^4b^2+b^2c^4+b^4c^2+c^2a^4+c^4a^2}\)

Đặt a²=x, b²=y, c²=z

=>\(VT^2\ge\frac{1}{8}.\frac{\left(x+y+z\right)^3}{x^2y+xy^2+y^2z+y^2z+z^2x+zx^2}\)(1)

Theo bđt Schur, ta có:

\(x\left(x-y\right)\left(x-z\right)+y\left(y-z\right)\left(y-x\right)+z\left(z-x\right)\left(z-y\right)\ge0\)

<=>\(x^3+y^3+z^3+3xyz\ge x^2y+xy^2+y^2z+y^2z+z^2x+zx^2\)

<=>\(x^3+y^3+z^3+6xyz+3\left(x^2y+xy^2+y^2z+y^2z+z^2x+zx^2\right)\ge4.\left(x^2y+xy^2+y^2z+y^2z+z^2x+zx^2\right)+3xyz\)

Vì \(xyz=\left(abc\right)^2\ge0\)

=>\(\left(x+y+z\right)^3\ge4\left(x^2y+xy^2+y^2z+y^2z+z^2x+zx^2\right)\)

=>\(\frac{\left(x+y+z\right)^3}{x^2y+xy^2+y^2z+y^2z+z^2x+zx^2}\ge4\)

Thay vào (1)=>\(VT^2\ge\frac{1}{2}=>VT\ge\frac{1}{\sqrt{2}}\)

=>ĐPCM

a,b,c>=0 mới được nhé

Đặt biểu thức là A

\(\sqrt{\frac{ab}{a^2+b^2}}=\frac{\sqrt{ab\left(a^2+b^2\right)}}{a^2+b^2}>=\frac{\sqrt{2abab}}{a^2}=\frac{\sqrt{2}ab}{a^2+b^2}\)

Dấu = xảy ra khi có một trong 2 số a,b =0 hoặc a=b.

Tương tự=> A>=\(\frac{\sqrt{2}ab}{a^2+b^2}+\frac{\sqrt{2}bc}{b^2+c^2}+\frac{\sqrt{2}ca}{a^2+c^2}\)

\(\sqrt{2}A>=\frac{2ab}{a^2+b^2}+\frac{2bc}{b^2+c^2}+\frac{2ca}{c^2+a^2}\)

\(\sqrt{2}A+3>=\frac{\left(a+b\right)^2}{a^2+b^2}+\frac{\left(b+c\right)^2}{b^2+c^2}+\frac{\left(c+a\right)^2}{c^2+a^2}.\)

>=\(\frac{\left(2a+2b+2c\right)^2}{2\left(a^2+b^2+c^2\right)}=\frac{4\left(a+b+c\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}=4.\)

=>A>=1/căn 2

Dấu = xảy ra khi 2 số bằng nhau, một số =0

xài mincopski thử, tui ăn cơm đã

#: Lỡ hẹn với Mincopxki rồi xài cách khác vậy :(

Đặt \(a=\frac{2x}{3};b=\frac{2y}{3};c=\frac{2z}{3}\)

Khi đó ta có \(xy+yz+xz\ge3\) và cần chứng minh

\(Σ_{cyc}\sqrt{\frac{4x^2}{9}+\frac{9}{\left(2y+3\right)^2}}\ge\frac{\sqrt{181}}{5}\)

Áp dụng BĐT Cauchy-Schwarz ta có:\(Σ_{cyc}\sqrt{\frac{4x^2}{9}+\frac{9}{\left(2y+3\right)^2}}\)

\(=\frac{15}{\sqrt{181}}Σ_{cyc}\sqrt{\left(\frac{4}{9}+\frac{9}{25}\right)\left(\frac{4x^2}{9}+\frac{9}{\left(2y+3\right)^2}\right)}\ge\frac{15}{\sqrt{181}}Σ_{cyc}\left(\frac{4x}{9}+\frac{9}{5\left(2y+3\right)}\right)\)

Giờ chỉ cần chứng minh \(\frac{15}{\sqrt{181}}Σ_{cyc}\left(\frac{4x}{9}+\frac{9}{5\left(2y+3\right)}\right)\ge\frac{\sqrt{181}}{5}\)

\(\Leftrightarrow20\left(x+y+z\right)+81\left(\frac{1}{2x+3}+\frac{1}{2y+3}+\frac{1}{2z+3}\right)\ge\frac{543}{5}\)

Đặt tiếp \(x+y+z=3u;xy+yz+xz=3v^2\left(v>0\right)\)

Vì thế \(u\ge v\ge1\)và áp dụng BĐT C-S dạng Engel ta có:

\(20\left(x+y+z\right)+81\left(\frac{1}{2x+3}+\frac{1}{2y+3}+\frac{1}{2z+3}\right)-\frac{543}{5}\)

\(\ge20\left(x+y+z\right)+81\cdot\frac{\left(1+1+1\right)^2}{Σ_{cyc}\left(2x+3\right)}-\frac{543}{5}=60u+\frac{729}{6u+9}-\frac{543}{5}\)

\(=3\left(20u+\frac{81}{2u+3}-\frac{181}{5}\right)=\frac{6\left(u-1\right)\left(100u+69\right)}{5\left(2u+3\right)}\ge0\) 

Điều này đúng tức là ta có ĐPCM

Bài 1:

\(BDT\Leftrightarrow\sqrt{\frac{3}{a+2b}}+\sqrt{\frac{3}{b+2c}}+\sqrt{\frac{3}{c+2a}}\le\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\)

\(\Leftrightarrow\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\ge\sqrt{3}\left(\frac{1}{\sqrt{a+2b}}+\frac{1}{\sqrt{b+2c}}+\frac{1}{\sqrt{c+2a}}\right)\)

Áp dụng BĐT Cauchy-Schwarz và BĐT AM-GM ta có: 

\(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{b}}\ge\frac{9}{\sqrt{a}+\sqrt{2}\cdot\sqrt{2b}}\ge\frac{9}{\sqrt{\left(1+2\right)\left(a+2b\right)}}=\frac{3\sqrt{3}}{\sqrt{a+2b}}\)

Tương tự cho 2 BĐT còn lại ta cũng có: 

\(\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}+\frac{1}{\sqrt{c}}\ge\frac{3\sqrt{3}}{\sqrt{b+2c}};\frac{1}{\sqrt{c}}+\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{a}}\ge\frac{3\sqrt{3}}{\sqrt{c+2a}}\)

Cộng theo vế 3 BĐT trên ta có: 

\(3\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)\ge3\sqrt{3}\left(\frac{1}{\sqrt{a+2b}}+\frac{1}{\sqrt{b+2c}}+\frac{1}{\sqrt{c+2a}}\right)\)

\(\Leftrightarrow\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\ge\sqrt{3}\left(\frac{1}{\sqrt{a+2b}}+\frac{1}{\sqrt{b+2c}}+\frac{1}{\sqrt{c+2a}}\right)\)

Đẳng thức xảy ra khi \(a=b=c\)

Bài 2: làm mãi ko ra hình như đề sai, thử a=1/2;b=4;c=1/2

Bài 2/

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

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

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

\(\ge\frac{\left(bc+ca+ab\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\)

\(\ge\frac{3\sqrt[3]{ab.bc.ca}}{2}=\frac{3}{2}\)

Dấu =  xảy ra khi \(a=b=c=1\)

bạn alibaba dòng thứ nhất rồi sao ra được dòng thứ hai á bạn mình k hiểu

1.

Áp dụng BĐT \(x^2+y^2+z^2\ge xy+yz+zx\)

\(\Rightarrow\left(\sqrt{ab}\right)^2+\left(\sqrt{bc}\right)^2+\left(\sqrt{ca}\right)^2\ge\sqrt{ab}.\sqrt{bc}+\sqrt{ab}.\sqrt{ac}+\sqrt{bc}.\sqrt{ac}\)

\(\Rightarrow ab+bc+ca\ge\sqrt{abc}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)\)

2.

\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt[]{\frac{ab.bc}{ca}}=2b\) ; \(\frac{ab}{c}+\frac{ac}{b}\ge2a\) ; \(\frac{bc}{a}+\frac{ac}{b}\ge2c\)

Cộng vế với vế:

\(2\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\right)\ge2\left(a+b+c\right)\)

\(\Leftrightarrow\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c\)

3.

Từ câu b, thay \(c=1\) ta được:

\(ab+\frac{b}{a}+\frac{a}{b}\ge a+b+1\)

4.

\(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}=\frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ac}\ge\frac{\left(a^2+b^2+c^2\right)}{ab+bc+ca}\)

\(\Rightarrow\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\ge\frac{\left(ab+bc+ca\right)^2}{ab+bc+ca}=ab+bc+ca\)

Dấu "=" xảy ra khi \(a=b=c\)

5.

\(\frac{a}{bc}+\frac{b}{ca}\ge2\sqrt{\frac{ab}{bc.ca}}=\frac{2}{c}\) ; \(\frac{a}{bc}+\frac{c}{ab}\ge\frac{2}{b}\) ; \(\frac{b}{ca}+\frac{c}{ab}\ge\frac{2}{a}\)

Cộng vế với vế:

\(2\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(\Rightarrow\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

1. bđt được viết lại thành

\(ab+bc+ca\ge a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}\)

Theo bđt AM-GM thì :

\(ab+bc\ge2\sqrt{ab\cdot bc}=2\sqrt{ab^2c}=2b\sqrt{ac}\)

Tương tự : \(bc+ca\ge2c\sqrt{ab}\); \(ab+ca\ge2a\sqrt{bc}\)

Cộng vế với vế

=> \(2\left(ab+bc+ca\right)\ge2\left(a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}\right)\)

=> \(ab+bc+ca\ge a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}\)( đpcm )

Dấu "=" xảy ra <=> a=b=c

https://olm.vn/thanhvien/chibiverycute là con chó

\(L.H.S=\Sigma_{cyc}\frac{a^2}{b}=\Sigma_{cyc}\left(\frac{a^2}{b}-a+b\right)=\Sigma_{cyc}\frac{a^2-ab+b^2}{b}\)

\(=\Sigma_{cyc}\left(\frac{a^2-ab+b^2}{b}+b\right)-\left(a+b+c\right)\)

\(\ge2\Sigma_{cyc}\sqrt{a^2-ab+b^2}-\left(a+b+c\right)\)

\(=\Sigma_{cyc}\sqrt{a^2-ab+b^2}+\Sigma_{cyc}\sqrt{\frac{1}{4}\left(a+b\right)^2+\frac{3}{4}\left(a-b\right)^2}-\left(a+b+c\right)\)

\(\ge\Sigma_{cyc}\sqrt{a^2-ab+b^2}+\Sigma_{cyc}\sqrt{\frac{1}{4}\left(a+b\right)^2}-\left(a+b+c\right)=\Sigma_{cyc}\sqrt{a^2-ab+b^2}=R.H.S\)

Đẳng thức xảy ra khi a = b = c

SOS ảo diệu hơn!

\(VT-VP=\Sigma_{cyc}\left[\frac{a^2}{b^2\left(\sqrt{\frac{a^2-ab+b^2}{b}}+\sqrt{b}\right)^2}^2+\Sigma_{cyc}\frac{3}{4\sqrt{a^2-ab+b^2}+2\left(a+b\right)}\right]\left(a-b\right)^2\ge0\) (đúng)

P/s: Nếu rảnh thì check hộ em:P

Bài 1: Ta có \(\left(\frac{a^2}{b}-a+b\right)+b^2=\frac{a^2-ab+b^2}{b}+b\ge2\sqrt{a^2-ab+b^2}\)  (áp dụng Bất Đẳng Thức Cosi)

\(=\sqrt{a^2-ab+b^2}+\sqrt{\frac{3}{4}\left(a-b\right)^2+\frac{1}{4}\left(a+b\right)^2}\ge\sqrt{a^2-ab+b^2}+\frac{1}{2}\left(a+b\right)\)

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

Tương tự ta có \(\hept{\begin{cases}\frac{b^2}{c}-b+2c\ge\sqrt{b^2-bc+c^2}+\frac{1}{2}\left(b+c\right)\left(2\right)\\\frac{c^2}{a}-c+2a\ge\sqrt{c^2-ac+a^2}+\frac{1}{2}\left(a+c\right)\left(3\right)\end{cases}}\)

Từ (1) và (2) và (3) \(\Rightarrow\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\sqrt{a^2-ab+b^2}+\sqrt{b^2-bc+c^2}+\sqrt{c^2-ac+a^2}\)

Dấu "=" xảy ra khi a=b=c