Chứng minh gì bạn?

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: \(\frac{x\sqrt{x}}{x+\sqrt{xy}+y}+\frac{y\sqrt{y}}{y+\sqrt{yz}+z}+\frac{z\sqrt{z}}{z+\sqrt{zx}+x}\ge\frac{2}{3}\).

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

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

Cộng vế với vế \(\Rightarrow VT=\dfrac{x-y}{\left(x+z\right)\left(y+z\right)}+\dfrac{y-z}{\left(x+y\right)\left(x+z\right)}+\dfrac{z-x}{\left(x+y\right)\left(y+z\right)}\)

\(=\dfrac{\left(x-y\right)\left(x+y\right)+\left(y-z\right)\left(y+z\right)+\left(z-x\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)

\(=\dfrac{x^2-y^2+y^2-z^2+z^2-x^2}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=0\)(đpcm)

Có \(VT=\dfrac{x^2}{x^3-xyz+2013x}+\dfrac{y^2}{y^3-xyz+2013y}+\dfrac{z^2}{z^3-xyz+2013z}\)

\(\ge\dfrac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2013\left(x+y+z\right)}\)

\(=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left[x^2+y^2+z^2-\left(xy+yz+zx\right)\right]+2013\left(x+y+z\right)}\)

\(=\dfrac{x+y+z}{x^2+y^2+z^2-\left(xy+yz+zx\right)+3\left(xy+yz+zx\right)}\) 

(vì \(2013=3.671=3\left(xy+yz+zx\right)\))

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

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

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

ĐTXR \(\Leftrightarrow\dfrac{1}{x^2-yz+2013}=\dfrac{1}{y^2-zx+2013}=\dfrac{1}{z^2-xy+2013}\)

\(\Leftrightarrow x^2-yz=y^2-zx=z^2-xy\)

\(\Leftrightarrow x=y=z\) (với \(x,y,z>0\))

Vậy ta có đpcm.

ta có bđt phụ ,,,,,,,,  x²+y²+z² >= xy+yz+zx

thay vào thôi,,,cái bđt dễ cm mà,,,nhân 2 2 vế rồi dùng tương đương

Đặt vế trái của BĐT cần chứng minh là P

Ta có:

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

\(P\ge\dfrac{\sqrt{\left(\sqrt{xy}+z\right)^2}+\sqrt{\left(x+y\right)^2}}{1+\sqrt{xy}}=\dfrac{\sqrt{xy}+x+y+z}{1+\sqrt{xy}}=\dfrac{\sqrt{xy}+1}{1+\sqrt{xy}}=1\) (đpcm)

Dấu "=" xảy ra khi \(x=y\)