Lời giải:

Giả sử $x>0; y< 0$. Khi đó:

\((xy-x^2)\sqrt{\frac{-y}{x}}=(y-x)x\sqrt{\frac{-y}{x}}=(y-x)\sqrt{-xy}\)

\((xy-y^2)\sqrt{\frac{-x}{y}}=(x-y)y\sqrt{\frac{-x}{y}}=(y-x)(-y)\sqrt{\frac{-x}{y}}=(y-x)\sqrt{(-y)^2.\frac{-x}{y}}=(y-x)\sqrt{-xy}\)

\(\Rightarrow (xy-x^2)\sqrt{\frac{-y}{x}}=(xy-y^2)\sqrt{\frac{-x}{y}}\Rightarrow \frac{xy-x^2}{\sqrt{\frac{-x}{y}}}=\frac{xy-y^2}{\sqrt{\frac{-y}{x}}}\) (đpcm)

1: \(A=\dfrac{x-2\sqrt{xy}+y}{x-y}=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)

2: Thay \(x=3+2\sqrt{2}\) và \(y=3-2\sqrt{2}\) vào A, ta được:

\(A=\dfrac{\sqrt{2}+1-\sqrt{2}+1}{\sqrt{2}+1+\sqrt{2}-1}=\dfrac{2}{2\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)

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

\(A=\sqrt{\dfrac{x^2}{x^2+\dfrac{1}{4}xy+y^2}}+\sqrt{\dfrac{y^2}{y^2+\dfrac{1}{4}yz+z^2}}+\sqrt{\dfrac{z^2}{z^2+\dfrac{1}{4}zx+x^2}}\le2\)

\(\Leftrightarrow\sqrt{\dfrac{1}{1+\dfrac{y}{4x}+\dfrac{y^2}{x^2}}}+\sqrt{\dfrac{1}{1+\dfrac{z}{4y}+\dfrac{z^2}{y^2}}}+\sqrt{\dfrac{1}{1+\dfrac{x}{4z}+\dfrac{x^2}{z^2}}}\le2\)

Đặt \(\left\{{}\begin{matrix}\dfrac{y}{x}=a\\\dfrac{z}{y}=b\\\dfrac{x}{z}=c\end{matrix}\right.\) thì bài toán thành

Chứng minh: \(A=\dfrac{1}{\sqrt{4a^2+a+4}}+\dfrac{1}{\sqrt{4b^2+b+4}}+\dfrac{1}{\sqrt{4c^2+c+4}}\le1\) với \(abc=1\)

Thử giải bài toán mới này xem sao bác.

*C/m bài toán mới của HUngnguyen

Ta có BĐT phụ \(\dfrac{1}{\sqrt{4a^2+a+4}}\le\dfrac{a+1}{2\left(a^2+a+1\right)}\)

\(\Leftrightarrow\left(a+1\right)^2\left(4a^2+a+4\right)\ge4\left(a^2+a+1\right)^2\)

\(\Leftrightarrow a\left(a-1\right)^2\ge0\)

Tương tự cho 2 BĐT còn lại cũng có:

\(\dfrac{1}{\sqrt{4b^2+b+4}}\le\dfrac{b+1}{2\left(b^2+b+1\right)};\dfrac{1}{\sqrt{4c^2+c+4}}\le\dfrac{c+1}{2\left(c^2+c+1\right)}\)

CỘng theo vế 3 BĐT trên ta có;

\(VT\le1=VP\) * Chỗ này tự giải chi tiết ra nhé, giờ bận rồi*

Bài này công kềnh vậy thôi thực ra nhìn cái là ra nó là hệ quả của BĐT Vasc của cụ Vasile Bat dang thuc Vasc.pdf

1.

\(\sqrt{\dfrac{x-1+\sqrt{2x-3}}{x+2-\sqrt{2x+3}}}\Leftrightarrow\)\(\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\sqrt{\dfrac{\left(\sqrt{2x-3}+1\right)^2}{\left(\sqrt{2x+3}-1\right)^2}}\end{matrix}\right.\)\(\Leftrightarrow\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{2x-3}+1}{\sqrt{2x+3}-1}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\left(\sqrt{2x-3}+1\right)\left(\sqrt{2x+3}+1\right)}{2\left(x+1\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\\dfrac{\sqrt{4x^2-9}+\sqrt{2x-3}+\sqrt{2x+3}+1}{2\left(x+1\right)}\end{matrix}\right.\)

hết tối giải rồi

Lời giải:

Ta có:

\(\frac{x}{\sqrt{x^2+1}}+\frac{y}{\sqrt{y^2+1}}+\frac{z}{\sqrt{z^2+1}}\)

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

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

Áp dụng BĐT Cauchy:

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

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

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

Cộng theo vế:

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

Ta có đpcm

Dấu bằng xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)

a:

Sửa đề: \(A=\left(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}+\dfrac{\sqrt{x^3}-\sqrt{y^3}}{y-x}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)

 \(A=\left(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}+\dfrac{x\sqrt{x}-y\sqrt{y}}{y-x}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)

\(A=\left(\sqrt{x}+\sqrt{y}-\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)

\(=\dfrac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)

\(=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)

b: căn xy>0

\(x-\sqrt{xy}+y=x-2\cdot\sqrt{x}\cdot\dfrac{1}{2}\sqrt{y}+\dfrac{1}{4}y+\dfrac{3}{4}y\)

\(=\left(\sqrt{x}-\dfrac{1}{2}\sqrt{y}\right)^2+\dfrac{3}{4}y>0\)

=>A>0

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 \(x=y\)