lẽ ra x,y,z>0 chứ sao lại a,b,c>0 :))

Áp dụng bđt Cô-si:\(x^2+yz\ge2\sqrt{x^2.yz}=2x\sqrt{yz}\Leftrightarrow\frac{1}{x^2+yz}\le\frac{1}{2x\sqrt{yz}}\)

tương tự: \(\frac{1}{y^2+xz}\le\frac{1}{2y\sqrt{xz}};\frac{1}{z^2+xy}\le\frac{1}{2z\sqrt{xy}}\)

=>\(\frac{1}{x^2+yz}\)\(+\frac{1}{y^2+xz}+\frac{1}{z^2+xy}\le\frac{1}{2x\sqrt{yz}}+\frac{1}{2y\sqrt{xz}}+\frac{1}{2z\sqrt{xy}}=\frac{\sqrt{xy}+\sqrt{yz}+\sqrt{xz}}{2xyz}\)

Mặt khác theo bđt Cô-si thì: \(\sqrt{xy}\le\frac{x+y}{2};\sqrt{yz}\le\frac{y+z}{2};\sqrt{xz}\le\frac{x+z}{2}\)

=>\(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\le\frac{x+y}{2}+\frac{y+z}{2}+\frac{x+z}{2}=\frac{2\left(x+y+z\right)}{2}=x+y+z\)

=>​\(\frac{1}{x^2+yz}+\frac{1}{y^2+xz}+\frac{1}{z^2+xy}\le\frac{\sqrt{xy}+\sqrt{yz}+\sqrt{xz}}{2xyz}\le\frac{x+y+z}{2xyz}=\frac{1}{2}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)\)

ta có đpcm.

Áp dụng cauchy cho mỗi mẫu số vế trái , có :

\(VT\le\frac{1}{2x\sqrt{yz}}+\frac{1}{2y\sqrt{xz}}+\frac{1}{2z\sqrt{xy}}=\frac{1}{2}.\left(\frac{1}{x\sqrt{yz}}+\frac{1}{y\sqrt{xz}}+\frac{1}{z\sqrt{xy}}\right)\)

                                         \(=\frac{1}{2}.\left(\frac{\sqrt{yz}}{xyz}+\frac{\sqrt{xz}}{xyz}+\frac{\sqrt{zx}}{xyz}\right)=\frac{1}{2}.\frac{\sqrt{yz}+\sqrt{xz}+\sqrt{xz}}{xyz}\)

Biến đổi vế phải , có :

\(VP=\frac{1}{2}.\left(\frac{z}{xyz}+\frac{x}{xyz}+\frac{y}{xyz}\right)=\frac{1}{2}.\frac{x+y+z}{xyz}\)

Ta có :

\(x+y+z\ge\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)

<=> \(2x+2y+2z\ge2\sqrt{xy}+2\sqrt{yz}+2\sqrt{zx}\) (đúng - Hệ quả của Cauchy, lên mạng sợt là ra )

=> \(\frac{1}{2}.\frac{x+y+z}{xyz}\ge\frac{1}{2}.\frac{\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}{xyz}\)

=> \(VP\ge VT\)

http://diendantoanhoc.net/topic/160455-%C4%91%E1%BB%81-to%C3%A1n-v%C3%B2ng-2-tuy%E1%BB%83n-sinh-10-chuy%C3%AAn-b%C3%ACnh-thu%E1%BA%ADn-2016-2017/

bài của tui mà -_-

hihi k biết làm nên đăng ^^

Theo bài ra ta có: \(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=1\Rightarrow x+y+z=xyz\)

Do:\(\sqrt{yz\left(1+x^2\right)}=\sqrt{yz+x^2yz}=\sqrt{yz+x\left(x+y+z\right)}=\sqrt{\left(x+y\right)\left(x+z\right)}\)

Tương tự: \(\sqrt{xy\left(1+z^2\right)}=\sqrt{\left(z+y\right)\left(x+z\right)}\);

\(\sqrt{zx\left(1+y^2\right)}=\sqrt{\left(z+y\right)\left(x+y\right)}\)

\(A=\sqrt{\frac{x^2}{yz\left(1+x^2\right)}}+\sqrt{\frac{y^2}{zx\left(1+y^2\right)}}+\sqrt{\frac{z^2}{xy\left(1+z^2\right)}}\)

\(A=\sqrt{\frac{x}{x+y}.\frac{x}{x+z}}+\sqrt{\frac{y}{x+y}.\frac{y}{y+z}}+\sqrt{\frac{z}{x+z}.\frac{z}{y+z}}\)

Áp dụng bất đẳng thức Cô si \(\frac{a+b}{2}\ge\sqrt{ab}\), dấu "=" xảy ra khi \(a=b\)

Ta có \(\sqrt{\frac{x}{x+y}.\frac{x}{x+z}}\le\frac{1}{2}\left(\frac{x}{x+y}+\frac{x}{x+z}\right)\);

\(\sqrt{\frac{y}{x+y}.\frac{y}{y+z}}\le\frac{1}{2}\left(\frac{y}{x+y}+\frac{y}{y+z}\right)\);

\(\sqrt{\frac{z}{x+z}.\frac{z}{y+z}}\le\frac{1}{2}\left(\frac{z}{x+z}+\frac{z}{y+z}\right)\)

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

Vậy \(A\le\frac{3}{2}\). Dấu "=" xảy ra khi \(x=y=z=\sqrt{3}\)

Bài 1: Theo đề : \(2ab+6bc+2ac=7abc\) \(;a,b,c>0\)

Chia cả 2 vế cho \(abc>0\Rightarrow\frac{2}{c}+\frac{6}{a}+\frac{2}{b}=7\)

Đặt: \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\Rightarrow\hept{\begin{cases}x,y,z>0\\2z+6x+2y=7\end{cases}}\)

Khi đó: \(M=\frac{4ab}{a+2b}+\frac{9ac}{a+4c}+\frac{4bc}{b+c}=\frac{4}{2x+y}+\frac{9}{4x+z}+\frac{4}{y+z}\)

\(\Rightarrow M=\frac{4}{2x+y}+2x+y+\frac{9}{4x+z}+4x+z+\frac{4}{y+z}+y+z-\left(2x+y+4x+z+y+z\right)\)

\(=\left(\frac{2}{\sqrt{x+2y}}-\sqrt{x+2y}\right)^2+\left(\frac{3}{\sqrt{4x+z}}-\sqrt{4x+z}\right)^2+\left(\frac{2}{\sqrt{y+z}}-\sqrt{y+z}\right)^2+17\ge17\)

Khi: \(\hept{\begin{cases}x=\frac{1}{2}\\y=z=1\end{cases}}\Rightarrow M=17\)

\(Min_M=17\Leftrightarrow a=2;b=1;c=1\)

ミ★๖ۣۜBăηɠ ๖ۣۜBăηɠ ★彡 chém bài khó nhất rồi nên em xin mạn phép chém bài dễ ạ.

2/\(VT=\Sigma_{cyc}\frac{\left(x+y+z\right)^2-x^2}{x\left(x+y+z\right)+yz}=\Sigma_{cyc}\frac{\left(y+z\right)\left(2x+y+z\right)}{\left(x+y\right)\left(x+z\right)}\)

\(\ge\Sigma_{cyc}\frac{\left(y+z\right)\left(2x+y+z\right)}{\frac{\left(2x+y+z\right)^2}{4}}=\Sigma_{cyc}\frac{4\left(y+z\right)}{2x+y+z}=\Sigma_{cyc}\frac{2\left(y+z-2x\right)}{2x+y+z}+6\)

\(=\Sigma_{cyc}\left(\frac{2\left(x+y+z\right)\left(y+z-2x\right)}{2x+y+z}-\frac{3}{2}\left(y+z-2x\right)\right)+6\)

\(=\Sigma_{cyc}\frac{\left(y+z-2x\right)^2}{2\left(2x+y+z\right)}+6\ge6\)

Cách 2 cho bài 2(suy ra từ cách 1). 

Với mọi x, y, z thỏa mãn \(x+y+z=1\) thì:

\(VT-VP=\Sigma_{cyc}\frac{\left(y-z\right)^2\left(y+z\right)}{\left(x+y\right)\left(x+z\right)\left(2x+y+z\right)}+\Sigma_{cyc}\frac{\left(y+z-2x\right)^2}{2\left(2x+y+z\right)}\ge0\)

@Nguyễn Việt Lâm

Bunhiacopxki: \(\left(x^2+yz+zx\right)\left(y^2+yz+zx\right)\ge\left(xy+yz+zx\right)^2\)

\(\Rightarrow\frac{xy}{x^2+yz+zx}\le\frac{xy\left(y^2+yz+zx\right)}{\left(xy+yz+zx\right)^2}\)

Thiết lập tương tự và cộng lại:

\(\Rightarrow VT\le\frac{xy\left(y^2+yz+zx\right)+yz\left(z^2+xy+zx\right)+zx\left(x^2+yz+xy\right)}{\left(xy+yz+zx\right)^2}\)

\(VT\le\frac{xy^3+xy^2z+x^2yz+yz^3+xy^2z+xyz^2+x^3z+xyz^2+x^2yz}{\left(xy+yz+zx\right)^2}\)

Ta chỉ cần chứng minh: \(\frac{xy^3+xy^2z+x^2yz+yz^3+xy^2z+xyz^2+x^3z+xyz^2+x^2yz}{\left(xy+yz+zx\right)^2}\le\frac{x^2+y^2+z^2}{xy+yz+zx}\)

\(\Leftrightarrow xy^3+xy^2z+x^2yz+yz^3+xy^2z+xyz^2+x^3z+xyz^2+x^2yz\le\left(x^2+y^2+z^2\right)\left(xy+yz+zx\right)\)

\(\Leftrightarrow x^2yz+xy^2z+xyz^2\le x^3y+y^3z+z^3x\)

\(\Leftrightarrow\frac{x^2}{z}+\frac{y^2}{x}+\frac{z^2}{y}\ge x+y+z\) (đúng theo Cauchy-Schwarz)

Dấu "=" xảy ra khi \(x=y=z\)

BĐT của bạn bị ngược dấu, mà có vẻ các mẫu số cũng ko đúng (để ý mẫu số thứ 2 và thứ 3 đều có chung xy+xz ko hợp lý)

Ta có \(\frac{x+2xy+1}{x+xy+xz+1}=\frac{x+2xy+xyz}{x+xy+xz+xyz}=\frac{1+2y+yz}{\left(y+1\right)\left(z+1\right)}\)

Tương tự => \(M=\frac{1+2y+yz}{\left(y+1\right)\left(z+1\right)}+\frac{1+2z+zx}{\left(1+x\right)\left(z+1\right)}+\frac{1+2x+xy}{\left(1+x\right)\left(y+1\right)}\)

=> \(M=\frac{\left(1+2y+yz\right)\left(1+x\right)+\left(1+2z+zx\right)\left(1+y\right)+\left(1+2x+xy\right)\left(1+z\right)}{\left(1+x\right)\left(1+y\right)\left(1+z\right)}\)

=>\(M=\frac{6+3\left(x+y+z\right)+3\left(xy+yz+xz\right)}{2+\left(x+y+z\right)+\left(xy+yz+xz\right)}=3\)