\(T\ge\dfrac{\left(x+y+z\right)^2}{x+y+z+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}\ge\dfrac{\left(x+y+z\right)^2}{x+y+z+x+y+z}=\dfrac{x+y+z}{2}\ge\dfrac{2019}{2}\)

áp dụng BĐT:\(\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\) với a,b,c,x,y,z là số dương

ta có BĐT Bunhiacopxki cho 3 bộ số:\(\left(\dfrac{a}{\sqrt{x}};\sqrt{x}\right);\left(\dfrac{b}{\sqrt{y}};\sqrt{y}\right);\left(\dfrac{c}{\sqrt{z}};\sqrt{z}\right)\)

ta có :

\(\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\left(x+y+z\right)\)\(=\left[\left(\dfrac{a}{\sqrt{x}}\right)^2+\left(\dfrac{b}{\sqrt{y}}\right)^2+\left(\dfrac{c}{\sqrt{z}}\right)^2\right]\).\(\left[\left(\sqrt{x}\right)^2+\left(\sqrt{y}\right)^2+\left(\sqrt{z}\right)^2\right]\)\(\ge\left(\dfrac{a}{\sqrt{x}}.\sqrt{x}+\dfrac{b}{\sqrt{y}}.\sqrt{y}+\dfrac{c}{\sqrt{z}}.\sqrt{z}\right)^2=\left(a+b+c\right)^2\)

lúc đó ta có :\(\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\ge\dfrac{\left(a+b+c\right)^2}{x+y+z}\)

ta có \(T=\dfrac{x^2}{x+\sqrt{yz}}+\dfrac{y^2}{y+\sqrt{zx}}+\dfrac{z^2}{z+\sqrt{xy}}\)\(\ge\dfrac{\left(x+y+z\right)^2}{x+\sqrt{yz}+y+\sqrt{zx}+z+\sqrt{xy}}\) mà ta có :

\(\sqrt{yz}+\sqrt{zx}+\sqrt{xy}\)\(\le\dfrac{x+y}{2}+\dfrac{x+z}{2}+\dfrac{z+y}{2}\)\(\Rightarrow\sqrt{yz}+\sqrt{zx}+\sqrt{xy}\le x+y+z\)

\(\Rightarrow T=\dfrac{2019}{2}\Leftrightarrow x=y=z=673\)

vậy \(\text{MinT}=\dfrac{2019}{2}\) khi và chỉ khi x=y=z=673

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

\(P=\dfrac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2x^2+xz+2z^2}+z\sqrt{2y^2+xy+2x^2}\right)\)

\(=\dfrac{1}{xyz}\left(x\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}+y\sqrt{\dfrac{5}{4}\left(x+z\right)^2+\dfrac{3}{4}\left(x-z\right)^2}+z\sqrt{\dfrac{5}{4}\left(x+y\right)^2+\dfrac{3}{4}\left(x-y\right)^2}\right)\)

\(\ge\dfrac{1}{xyz}\left[x.\dfrac{\sqrt{5}\left(z+y\right)}{2}+y.\dfrac{\sqrt{5}\left(x+z\right)}{2}+z.\dfrac{\sqrt{5}\left(x+y\right)}{2}\right]\)

\(=\dfrac{\sqrt{5}\left(z+y\right)}{2yz}+\dfrac{\sqrt{5}\left(x+z\right)}{2xz}+\dfrac{\sqrt{5}\left(x+y\right)}{2xy}\)

\(=\dfrac{\sqrt{5}}{3}\left(1+1+1\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge\dfrac{\sqrt{5}}{3}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2=\dfrac{\sqrt{5}}{3}\) (bunhia)

Dấu = xảy ra khi \(x=y=z=9\)

 Thấy : \(\sqrt{2y^2+yz+2z^2}=\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}\ge\dfrac{\sqrt{5}}{2}\left(y+z\right)>0\) 

CMTT : \(\sqrt{2x^2+xz+2z^2}\ge\dfrac{\sqrt{5}}{2}\left(x+z\right)\)  ; \(\sqrt{2y^2+xy+2x^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\) 

Suy ra : \(P\ge\dfrac{1}{xyz}.\dfrac{\sqrt{5}}{2}\left[x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\right]\)

\(\Rightarrow P\ge\sqrt{5}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\) 

Ta có : \(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}=\sqrt{xyz}\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\) 

Mặt khác :   \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2}{3}=\dfrac{1}{3}\)

Suy ra : \(P\ge\dfrac{\sqrt{5}}{3}\)

" = " \(\Leftrightarrow x=y=z=9\)

ta có \(\sqrt{x^2-xy+y^2}=\sqrt{\frac{1}{4}\left(x+y\right)^2+\frac{3}{4}\left(x-y\right)^2}\ge\sqrt{\frac{1}{4}\left(x+y\right)^2}=\frac{1}{2}\left(x+y\right)\)

tương tự ta có các trường hợp còn lại và ta có 

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

đặt \(x+y=a;y+z=b;z+x=c\)

=> \(S\ge\frac{1}{2}\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)\)

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

Áp dụng bđt svác sơ ta có 

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

mạt khác Áp dụng bđt cô si ta có 

\(\hept{\begin{cases}a^2+b^2\ge2ab\\b^2+c^2\ge2bc\\c^2+a^2\ge2ac\end{cases}}\)

=> \(a^2+b^2+c^2\ge2\left(ab+bc+ca\right)\)

=> \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

=> \(A\ge\frac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\frac{3}{2}\)

=> \(S\ge\frac{3}{4}\)

dấu = xảy ra <=> x=y=z>o

ta có \(\sqrt{x^2-xy+y^2}=\sqrt{\frac{1}{4}x^2+\frac{1}{2}xy+\frac{1}{4y^2}+\frac{3}{4}x^2-\frac{3}{2}xy+\frac{3}{4}y^2}\)

       \(=\sqrt{\frac{1}{4}\left(x^2+2xy+y^2\right)+\frac{3}{4}\left(x^2-2xy+y^2\right)}=\sqrt{\frac{1}{4}\left(x+y\right)^2+\frac{3}{4}\left(x-y\right)^2}\)

Ta có

          \(\sqrt{x^2-xy+y^2}=\sqrt{\frac{1}{4}\left(x+y\right)^2+\frac{3}{4}\left(x-y\right)^2}\ge\sqrt{\frac{1}{4}\left(x+y\right)^2}=\frac{1}{2}\cdot\left(y+x\right)\)

Tương tự ta có các TH còn lại

Ta có

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

Đặt x + y = a ; y + z = b ; z + x = c

    \(S\ge\frac{1}{2}\left(\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\right)\)

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

Áp dụng BĐT Svac sơ ta có

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

Mặt khác ADBĐT Cosy ta có

    \(\hept{\begin{cases}a^2+b^2\ge2ab\\b^2+c^2\ge2bc\\c^2+a\ge^22ac\end{cases}}\)

            \(\Rightarrow a^2+b^2+c^2\ge2\left(ab+bc+ca\right)\)

             \(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

             \(\Rightarrow A\ge\frac{3}{2}\Rightarrow S\ge\frac{3}{4}\)

Dấu =xảy ra khi x=y=z>o

\(\downarrow\)

Ta có : 2P = \(\frac{\sqrt{4x^2-4xy+4y^2}}{x+y+2z}+\frac{\sqrt{4y^2-4yz+4z^2}}{y+z+2x}+\frac{\sqrt{4z^2-4zx+4x^2}}{z+x+2y}\)

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

Lại có  \(\frac{\sqrt{\left[\left(2x-y\right)^2+\left(\sqrt{3}y\right)^2\right]\left[\left(1^2+\left(\sqrt{3}\right)^2\right)\right]}}{x+y+2z}\ge\frac{\left[\left(2x-y\right).1+3y\right]}{x+y+2z}=\frac{2\left(x+y\right)}{x+y+2z}\)

=> \(\sqrt{\frac{\left(2x-y\right)^2+\left(\sqrt{3}y\right)^2}{x+y+2z}}\ge\frac{x+y}{x+y+2z}\)(BĐT Bunyakovsky) 

Tương tự ta đươc \(2P\ge\frac{x+y}{x+y+2z}+\frac{y+z}{2x+y+z}+\frac{z+x}{2y+z+x}\)

Đặt x + y = a ; y + z = b ; x + z = c

Khi đó \(2P\ge\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\left(a+b+c\right)\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)-3\)

\(\ge\left(a+b+c\right).\frac{9}{2\left(a+b+c\right)}-3\ge\frac{9}{2}-3=\frac{3}{2}\)

=> \(P\ge\frac{3}{4}\)

Dấu "=" xảy ra <=> x = y = z 

bài 8 : bỏ dấu hoặc  rồi tính 

a;( 17 - 299) + ( 17 - 25 + 299)

bằng 20 ấn mtinh ra thế

Có \(\sqrt{\dfrac{xy}{x+y+2z}}=\dfrac{\sqrt{xy}}{\sqrt{x+y+2z}}\)\(=\dfrac{2\sqrt{xy}}{\sqrt{\left(1+1+2\right)\left(x+y+2z\right)}}\)\(\le\dfrac{2\sqrt{xy}}{\sqrt{x}+\sqrt{y}+2\sqrt{z}}\) (theo bunhia dưới mẫu)\(\le\dfrac{2\sqrt{xy}}{4}\left(\dfrac{1}{\sqrt{x}+\sqrt{z}}+\dfrac{1}{\sqrt{y}+\sqrt{z}}\right)\)

\(\Leftrightarrow\sqrt{\dfrac{xy}{x+y+2z}}\le\dfrac{1}{2}\left(\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{z}}+\dfrac{\sqrt{xy}}{\sqrt{y}+\sqrt{z}}\right)\)

Tương tự cũng có:

\(\sqrt{\dfrac{yz}{y+z+2x}}\le\dfrac{1}{2}\left(\dfrac{\sqrt{yz}}{\sqrt{y}+\sqrt{x}}+\dfrac{\sqrt{yz}}{\sqrt{z}+\sqrt{x}}\right)\)

\(\sqrt{\dfrac{zx}{z+x+2y}}\le\dfrac{1}{2}\left(\dfrac{\sqrt{zx}}{\sqrt{z}+\sqrt{y}}+\dfrac{\sqrt{zx}}{\sqrt{x}+\sqrt{y}}\right)\)

Cộng vế với vế ta được:

 \(VT\le\dfrac{1}{2}\left(\dfrac{\sqrt{xy}+\sqrt{yz}}{\sqrt{x}+\sqrt{z}}+\dfrac{\sqrt{xy}+\sqrt{zx}}{\sqrt{y}+\sqrt{z}}+\dfrac{\sqrt{yz}+\sqrt{zx}}{\sqrt{x}+\sqrt{y}}\right)\)

\(\Leftrightarrow VT\le\dfrac{1}{2}\left(\sqrt{y}+\sqrt{x}+\sqrt{z}\right)=\dfrac{1}{2}\)

Dấu = xảy ra khi \(x=y=z=\dfrac{1}{9}\)

hay

Ta có \(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\sqrt{xyz}\left(x,y,z>0\right)\).

\(\Leftrightarrow\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}=1\).

\(P=\frac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2z^2+xz+2x^2}+z\sqrt{2x^2+xy+y^2}\right)\)\(\left(x,y,z>0\right)\).

Ta có: 

\(\sqrt{2y^2+2yz+2z^2}=\sqrt{\frac{5}{4}\left(y^2+2yz+z^2\right)+\frac{3}{4}\left(y^2-2yz+z^2\right)}\)

\(=\sqrt{\frac{5}{4}\left(y+z\right)^2+\frac{3}{4}\left(y-z\right)^2}\).

Ta có:

\(\frac{3}{4}\left(y-z\right)^2\ge0\forall y;z>0\).

\(\Leftrightarrow\frac{3}{4}\left(y-z\right)^2+\frac{5}{4}\left(y+z\right)^2\ge\frac{5}{4}\left(y+z\right)^2\forall y;z>0\).

\(\Rightarrow\sqrt{\frac{3}{4}\left(y-z\right)^2+\frac{5}{4}\left(y+z\right)^2}\ge\frac{\sqrt{5}}{2}\left(y+z\right)\forall y,z>0\).

\(\Leftrightarrow\sqrt{2y^2+yz+2z^2}\ge\frac{\sqrt{5}}{2}\left(y+z\right)\forall y;z>0\).

\(\Leftrightarrow x\sqrt{2y^2+yz+2z^2}\ge\frac{\sqrt{5}}{2}x\left(y+z\right)\forall x;y;z>0\left(1\right)\).

Chứng minh tương tự, ta được:

\(y\sqrt{2x^2+xz+2z^2}\ge\frac{\sqrt{5}}{2}y\left(x+z\right)\forall x;y;z>0\left(2\right)\).

Chứng minh tương tự, ta được:

\(z\sqrt{2x^2+xy+2y^2}\ge\frac{\sqrt{5}}{2}z\left(x+y\right)\forall x;y;z>0\left(3\right)\).

Từ \(\left(1\right),\left(2\right),\left(3\right)\), ta được:

\(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2z^2+xz+2x^2}+z\sqrt{2x^2+xy+2y^2}\)\(\ge\)\(\frac{\sqrt{5}}{2}\left[x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\right]=\sqrt{5}\left(xy+yz+zx\right)\).

\(\Leftrightarrow\frac{1}{xyz}\left(x\sqrt{2y^2+yz+z^2}+y\sqrt{2z^2+zx+2x^2}+z\sqrt{2x^2+xy+2y^2}\right)\)\(\ge\)\(\frac{\sqrt{5}\left(xy+yz+zx\right)}{xyz}=\sqrt{5}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\).

\(\Leftrightarrow P\ge\frac{\sqrt{5}}{3}.3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{\sqrt{5}}{3}\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\)

\(\left(4\right)\).

Vì \(x,y,z>0\)nên áp dụng bất đẳng thức Bu-nhi-a-cốp-xki, ta được:
\(\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\ge\)\(\left(1.\frac{1}{\sqrt{x}}+1.\frac{1}{\sqrt{y}}+1.\frac{1}{\sqrt{z}}\right)^2\).

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

(vì\(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}=1\)).

\(\Leftrightarrow\frac{\sqrt{5}}{3}\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\ge\frac{\sqrt{5}}{3}\)\(\left(5\right)\).

Từ \(\left(4\right)\)và \(\left(5\right)\), ta được:

\(P\ge\frac{\sqrt{5}}{3}\).

Dấu bằng xảy ra.

\(\Leftrightarrow\hept{\begin{cases}x=y=z>0\\\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\sqrt{xyz}\end{cases}}\Leftrightarrow x=y=z=9\).

Vậy \(minP=\frac{\sqrt{5}}{3}\Leftrightarrow x=y=z=9\).

hình như thiếu cái gì đó

a) Giả sử \(x^2-xy+y^2\ge\frac{1}{3}\left(x^2+xy+y^2\right)\)

\(\Leftrightarrow3\left(x^2-xy+y^2\right)\ge\frac{1}{3}.3\left(x^2+xy+y^2\right)\)

\(\Leftrightarrow3\left(x^2-xy+y^2\right)\ge x^2+xy+y^2\)

\(\Leftrightarrow3x^2-3xy+3y^2-x^2-xy-y^2\ge0\)

\(\Leftrightarrow2x^2-4xy+2y^2\ge0\)

\(\Leftrightarrow2\left(x^2-2xy+y^2\right)\ge0\)

\(\Leftrightarrow2\left(x-y\right)^2\ge0\)(luôn đúng với mọi \(x,y\in R\)).

Dấu bằng xảy ra\(\Leftrightarrow x=y\).

Vậy \(x^2-xy+y^2\ge\frac{1}{3}\left(x^2+xy+y^2\right)\)với \(x,y\in R\).

Đặt \(A=\frac{x\sqrt{x}}{x+\sqrt{xy}+y}+\frac{y\sqrt{y}}{y+\sqrt{yz}+z}+\frac{z\sqrt{z}}{z+\sqrt{zx}+x}\left(x,y,z>0\right)\)

Và đặt \(B=\frac{y\sqrt{y}}{x+\sqrt{xy}+y}+\frac{z\sqrt{z}}{y+\sqrt{yz}+z}+\frac{x\sqrt{x}}{z+\sqrt{zx}+x}\left(x,y,z>0\right)\)

Đặt \(\sqrt{x}=m,\sqrt{y}=n,\sqrt{z}=p\left(m,n,p>0\right)\)thì theo đề bài : \(m+n+p=2\)

Lúc đó:

\(A=\frac{m^2.m}{m^2+mn+n^2}+\frac{n^2.n}{n^2+np+p^2}+\frac{p^2.p}{p^2+pm+m^2}\)

\(A=\frac{m^3}{m^2+mn+n^2}+\frac{n^3}{n^2+np+p^2}+\frac{p^3}{p^2+pm+m^2}\)

Và \(B=\frac{n^3}{m^2+mn+n^2}+\frac{p^3}{n^2+np+p^2}+\frac{m^3}{p^2+pm+m^2}\)

Xét hiệu \(A-B=\frac{m^3-n^3}{m^2+mn+n^2}+\frac{n^3-p^3}{n^2+np+p^2}+\frac{p^3-m^3}{p^2+pm+m^2}\)

\(\Leftrightarrow A-B=\frac{\left(m-n\right)\left(m^2+mn+n^2\right)}{m^2+mn+n^2}+\frac{\left(n-p\right)\left(n^2+np+p^2\right)}{n^2+np+p^2}\)\(+\frac{\left(p-m\right)\left(p^2+pm+m^2\right)}{p^2+pm+m^2}\)

\(\Leftrightarrow A-B=\left(m-n\right)+\left(n-p\right)+\left(p-m\right)\)

\(\Leftrightarrow A-B=m-n+n-p+p-m=0\)

\(\Leftrightarrow A=B\)

Xét \(A+B=\frac{m^3+n^3}{m^2+mn+n^2}+\frac{n^3+p^3}{n^2+np+p^2}+\frac{p^3+m^3}{p^2+pm+m^2}\)

\(\Leftrightarrow A+A=2A=\frac{\left(m+n\right)\left(m^2-mn+n^2\right)}{m^2+m+n^2}+\frac{\left(n+p\right)\left(n^2-np+p^2\right)}{n^2+np+p^2}\)\(\frac{\left(p+m\right)\left(p^2-pm+m^2\right)}{p^2+pm+m^2}\)

Theo câu a), ta có \(x^2-xy+y^2\ge\frac{1}{3}\left(x^2+xy+y^2\right)\)với \(x,y\in R\)

\(\Leftrightarrow\frac{x^2-xy+y^2}{x^2+xy+y^2}\ge\frac{1}{3}\left(1\right)\)

Dấu bằng xảy ra \(\Leftrightarrow x=y\)

Áp dụng bất đẳng thức (1) (với \(m,n>0\)), ta được:

\(\frac{m^2-mn+n^2}{m^2+mn+n^2}\ge\frac{1}{3}\)

\(\Leftrightarrow\frac{\left(m+n\right)\left(m^2-mn+n^2\right)}{m^2+mn+n^2}\ge\frac{m+n}{3}\left(2\right)\)

Dấu bằng xảy ra \(\Leftrightarrow m=n>0\)

Chứng minh tương tự, ta được:

\(\frac{\left(n+p\right)\left(n^2-np+p^2\right)}{n^2+np+p^2}\ge\frac{n+p}{3}\left(3\right)\)

Dấu bằng xảy ra\(\Leftrightarrow n=p>0\)

\(\frac{\left(p+m\right)\left(p^2-pm+m^2\right)}{p^2+pm+m^2}\ge\frac{p+m}{2}\left(4\right)\)

Dấu bằng xảy ra\(\Leftrightarrow p=m>0\)

Từ \(\left(2\right),\left(3\right),\left(4\right)\), ta được:

\(\frac{\left(m+n\right)\left(m^2-mn+n^2\right)}{m^2+mn+n^2}+\frac{\left(n+p\right)\left(n^2-np+p^2\right)}{n^2+np+p^2}\)\(+\frac{\left(p+m\right)\left(p^2-pm+m^2\right)}{p^2-pm+m^2}\ge\frac{m+n}{3}+\frac{n+p}{3}+\frac{p+m}{3}\)

\(\Leftrightarrow2A\ge\frac{m+n+n+p+p+m}{3}\)

\(\Leftrightarrow2A\ge\frac{2\left(m+n+p\right)}{3}\)

\(\Leftrightarrow A\ge\frac{m+n+p}{3}\)

\(\Leftrightarrow A\ge\frac{2}{3}\)(vì \(m+n+p=2\)) (điều phải chứng minh).

Dấu bằng xảy ra.

\(\Leftrightarrow\hept{\begin{cases}m=n=p>0\\m+n+p=2\end{cases}}\Leftrightarrow m=n=p=\frac{2}{3}\)\(\Leftrightarrow\sqrt{x}=\sqrt{y}=\sqrt{z}=\frac{2}{3}\Leftrightarrow x=y=z=\frac{4}{9}\)

Vậy nếu \(x,y,z>0\) và \(\sqrt{x}+\sqrt{y}+\sqrt{z}=2\)thì: \(\frac{x\sqrt{x}}{x+\sqrt{xy}+y}+\frac{y\sqrt{y}}{y+\sqrt{yz}+z}+\frac{z\sqrt{z}}{z+\sqrt{zx}+x}\ge\frac{2}{3}\).