Áp dụng BĐT BSC và BĐT Cosi:

\(17\left(x+y+z\right)+2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

\(\ge17\left(x+y+z\right)+\frac{2.\left(1+1+1\right)^2}{x+y+z}\)

\(=17\left(x+y+z\right)=\frac{18}{x+y+z}\)

\(=17\left(x+y+z\right)=\frac{17}{x+y+z}+\frac{1}{x+y+z}\)

\(\ge2\sqrt{17\left(x+y+z\right).\frac{17}{x+y+z}}+\frac{1}{1}\)

\(=35\)

\(\Rightarrow17\left(x+y+z\right)+2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge35\)

Đẳng thức xảy ra khi \(x=y=z=\frac{1}{3}\)

Áp dụng bất đẳng thức AM-GM kết hợp giả thiết x + y + z ≤ 1 ta có :

\(17\left(x+y+z\right)+2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=17x+17y+17z+\frac{2}{x}+\frac{2}{y}+\frac{2}{z}\)

\(=\left(18x+\frac{2}{x}\right)+\left(18y+\frac{2}{y}\right)+\left(18z+\frac{2}{z}\right)-\left(x+y+z\right)\)

\(\ge2\sqrt{18x\cdot\frac{2}{x}}+2\sqrt{18y\cdot\frac{2}{y}}+2\sqrt{18z\cdot\frac{2}{z}}-1=12\cdot3-1=35\)( đpcm )

Dấu "=" xảy ra <=> x=y=z=1/3

Ta có:

\(1+x^2=xy+yz+zx+x^2=\left(x+y\right)\left(x+z\right)\)

\(1+y^2=xy+yz+xz+y^2=\left(y+z\right)\left(x+y\right)\)

\(1+z^2=xy+yz+xz+z^2=\left(x+z\right)\left(y+z\right)\)

Thay vào A được:

\(P=x\sqrt{\frac{\left(y+z\right)\left(x+y\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}+y\sqrt{\frac{\left(x+z\right)\left(y+z\right)\left(x+y\right)\left(x+z\right)}{\left(y+z\right)\left(x+y\right)}}\)\(+z\sqrt{\frac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(x+y\right)}{\left(x+z\right)\left(y+z\right)}}\)

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

\(=x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\)

\(=xy+xz+xy+yz+xz+zy\)

\(=2\left(xy+yz+xz\right)\)

\(=2\)(do xy+yz+xz=1)

=>Đpcm

Dạng toán này rất nhiều bạn hỏi rồi: thay \(xy+yz+zx=1\) vào các căn thức rồi phân tích đa thức thành nhân tử.

\(\Sigma\dfrac{a^2}{\left(2a+b\right)\left(2a+c\right)}=\Sigma\left(\dfrac{1}{9}.\dfrac{a^2\left(2+1\right)^2}{2a.\left(\Sigma a\right)+2a^2+bc}\right)\le\Sigma\left(\dfrac{1}{9}.\dfrac{4a^2}{2a\left(\Sigma a\right)}+\dfrac{1}{9}.\dfrac{a^2}{2a^2+bc}\right)\)

\(=\Sigma\left(\dfrac{1}{9}.\left(\dfrac{2a}{\Sigma a}+\dfrac{a^2}{2a^2+bc}\right)\right)=\dfrac{1}{9}\left(2+\Sigma\dfrac{a^2}{2a^2+bc}\right)\)

Cần chứng minh \(\Sigma\frac{a^2}{2a^2+bc}\le1\)

<=> \(\Sigma\frac{bc}{2a^2+bc}\ge1\)         (*)

Đặt (x;y;z) ------->  \(\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\)

Suy ra (*)  <=>  \(\Sigma\frac{x^2}{x^2+2xy}\ge1\Leftrightarrow\frac{\Sigma x^2}{\Sigma x^2}\ge1\) (đúng)

Vậy \(\Sigma\frac{a^2}{2a^2+bc}\le1\)

Suy ra \(\Sigma\frac{a^2}{\left(2a+b\right)\left(2a+c\right)}\le\frac{1}{9}\left(2+\Sigma\frac{a^2}{2a^2+bc}\right)\le\frac{1}{9}\left(2+1\right)=\frac{1}{3}\)

Đẳng thức xảy ra <=> x = y = z = 1 

Nguồn : Trần Thắng

Theo AM - GM và Bunhiacopski ta có được 

\(x^2+y^2\ge\frac{\left(x+y\right)^2}{2};\frac{1}{x^2}+\frac{1}{y^2}\ge\frac{2}{xy}\ge\frac{8}{\left(x+y\right)^2}\)

Khi đó \(LHS\ge\left[\frac{\left(x+y\right)^2}{2}+z^2\right]\left[\frac{8}{\left(x+y\right)^2}+\frac{1}{z^2}\right]\)

\(\)\(=\left[\frac{1}{2}+\left(\frac{z}{x+y}\right)^2\right]\left[8+\left(\frac{x+y}{z}\right)^2\right]\)

Đặt \(t=\frac{z}{x+y}\ge1\)

Khi đó:\(LHS\ge\left(\frac{1}{2}+t^2\right)\left(8+\frac{1}{t^2}\right)=8t^2+\frac{1}{2t^2}+5\)

\(=\left(\frac{1}{2t^2}+\frac{t^2}{2}\right)+\frac{15t^2}{2}+5\ge\frac{27}{2}\)

Vậy ta có đpcm

Ta có:

\(VT-VP=\frac{\left(x^2+y^2\right)\left(\Sigma xy\right)\left(\Sigma x\right)\left[z\left(x+y\right)-xy\right]\left(z-x-y\right)}{x^2y^2z^2\left(x+y\right)^2}+\frac{\left(x-y\right)^2\left(2x+y\right)^2\left(x+2y\right)^2}{2x^2y^2\left(x+y\right)^2}\ge0\)

Vì \(z\left(x+y\right)-xy\ge\left(x+y\right)^2-xy\ge4xy-xy>0\) 

Đặt a=x-2; b=y-2; c=z-2. Phải chứng minh abc =<1

Thật vậy, từ \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)ta có:

\(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=1\)

Theo BĐT Cauchy ta có:

\(\frac{1}{a+2}=\left(\frac{1}{2}-\frac{1}{b+2}\right)+\left(\frac{1}{2}-\frac{1}{c+2}\right)=\frac{1}{2}\left(\frac{b}{b+2}+\frac{c}{c+2}\right)\ge\sqrt{\frac{bc}{\left(b+2\right)\left(c+2\right)}}\left(1\right)\)

tương tự \(\hept{\begin{cases}\frac{1}{b+2}\ge\sqrt{\frac{ac}{\left(a+2\right)\left(c+2\right)}}\left(2\right)\\\frac{1}{c+2}\ge\sqrt{\frac{ab}{\left(a+2\right)\left(b+2\right)}}\left(3\right)\end{cases}}\)

Nhân các vế của (1)(2)(3) ta được đpcm

Dấu "=" xảy ra <=> a=b=c hay x=y=z=3