Biến đổi tương đương:

\(\Leftrightarrow\dfrac{x^2+y^2}{xy}\ge2\)

\(\Leftrightarrow x^2+y^2\ge2xy\)

\(\Leftrightarrow x^2+y^2-2xy\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng)

Vậy BĐT đã được chứng minh

Cách khác so với anh Nguyễn Việt Lâm

Ta có: \(\dfrac{x}{y}+\dfrac{y}{x}\ge2\sqrt{\dfrac{x}{y}\cdot\dfrac{y}{x}}=2\)  (đpcm)

ÁP dụng AM-GM:

\(\sum\dfrac{a^2}{\sqrt{1-a^2}}=\sum\dfrac{a^3}{\sqrt{\left(1-a^2\right).a^2}}\ge\sum\dfrac{a^3}{\dfrac{1}{2}\left(1-a^2+a^2\right)}=2\sum a^3=2\left(đpcm\right)\)

Dấu = không xảy ra

Ta có:

\(VT=2+\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{y}{z}+\dfrac{x}{z}+\dfrac{z}{x}\)

Do đó ta chỉ cần chứng minh:

\(\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{z}{x}+\dfrac{x}{z}\ge\dfrac{2\left(x+y+z\right)}{\sqrt[3]{xyz}}\)

Ta có:

\(\dfrac{x}{y}+\dfrac{x}{y}+1\ge3\sqrt[3]{\dfrac{x^2}{y^2}}\) 

Tương tự ...

Cộng lại ta có:

\(2\left(\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{z}{x}+\dfrac{x}{z}\right)+6\ge3\left(\sqrt[3]{\dfrac{x^2}{y^2}}+\sqrt[3]{\dfrac{y^2}{x^2}}+\sqrt[3]{\dfrac{y^2}{z^2}}+\sqrt[3]{\dfrac{z^2}{y^2}}+\sqrt[3]{\dfrac{z^2}{x^2}}+\sqrt[3]{\dfrac{x^2}{z^2}}\right)\)

\(\Rightarrow\dfrac{x}{y}+\dfrac{y}{x}+\dfrac{y}{z}+\dfrac{z}{y}+\dfrac{z}{x}+\dfrac{x}{z}\ge\sqrt[3]{\dfrac{x^2}{y^2}}+\sqrt[3]{\dfrac{y^2}{x^2}}+\sqrt[3]{\dfrac{y^2}{z^2}}+\sqrt[3]{\dfrac{z^2}{y^2}}+\sqrt[3]{\dfrac{z^2}{x^2}}+\sqrt[3]{\dfrac{x^2}{z^2}}\)

Do đó ta chỉ cần chứng minh:

\(\sqrt[3]{\dfrac{x^2}{y^2}}+\sqrt[3]{\dfrac{y^2}{x^2}}+\sqrt[3]{\dfrac{y^2}{z^2}}+\sqrt[3]{\dfrac{z^2}{y^2}}+\sqrt[3]{\dfrac{z^2}{x^2}}+\sqrt[3]{\dfrac{x^2}{z^2}}\ge\dfrac{2\left(x+y+z\right)}{\sqrt[3]{xyz}}\)

\(\Leftrightarrow\left(\sqrt[3]{\dfrac{x}{y}}-\sqrt[3]{\dfrac{x}{z}}\right)^2+\left(\sqrt[3]{\dfrac{y}{x}}-\sqrt[3]{\dfrac{y}{z}}\right)^2+\left(\sqrt[3]{\dfrac{z}{x}}-\sqrt[3]{\dfrac{z}{y}}\right)^2\ge0\) (luôn đúng)

Áp dụng BĐT: \(\dfrac{a^n+b^n}{2}\ge\left(\dfrac{a+b}{2}\right)^n\Rightarrow a^n+b^2\ge2\left(\dfrac{a+b}{2}\right)^n\):

\(\left(1+\dfrac{x}{y}\right)^{2018}+\left(1+\dfrac{y}{x}\right)^{2018}\ge2\left(\dfrac{2+\dfrac{x}{y}+\dfrac{y}{x}}{2}\right)^{2018}\ge2\left(\dfrac{2+2\sqrt{\dfrac{x}{y}.\dfrac{y}{x}}}{2}\right)^{2018}=2^{2019}\)

Dấu "=" xảy ra khi \(x=y\)

Ta có \(\dfrac{1}{1+x}\ge1-\dfrac{1}{1+y}+1-\dfrac{1}{1+x}=\dfrac{y}{1+y}+\dfrac{z}{1+z}\)

\(\ge2\sqrt{\dfrac{yz}{\left(y+1\right)\left(z+1\right)}}\)

Chứng minh tương tự, ta có

\(\dfrac{1}{1+y}\ge2\sqrt{\dfrac{xz}{\left(z+1\right)\left(x+1\right)}};\dfrac{1}{1+z}\ge2\sqrt{\dfrac{xy}{\left(x+1\right)\left(y+1\right)}}\)

Nhân cả 3 cua 3 BĐT cùng chiều, ta có

\(\dfrac{1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\ge\dfrac{8xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\Rightarrow xuz\le\dfrac{1}{8}\left(ĐPCM\right)\)

Lời giải:

Gọi biểu thức đã cho là $P$. Đặt $\sqrt{xy}=a; \sqrt{yz}=b$ với $a,b>0$ thì ta cần chứng minh:

$P=\frac{a}{1+b}+\frac{1}{a+b}+\sqrt{\frac{2b}{a+1}}\geq 2$

Áp dụng BĐT AM-GM:

\(\frac{a+1}{2b}.1\leq \left(\frac{\frac{a+1}{2b}+1}{2}\right)^2=(\frac{a+1+2b}{4b})^2\)

\(\Rightarrow \sqrt{\frac{2b}{a+1}}\geq \frac{4b}{a+2b+1}(1)\)

Áp dụng BĐT Cauchy-Schwarz:

\(\frac{a}{1+b}+\frac{1}{a+b}=\frac{a+b+1}{b+1}+\frac{a+b+1}{a+b}-2=(a+b+1)(\frac{1}{b+1}+\frac{1}{a+b})-2\geq \frac{4(a+b+1)}{a+2b+1}-2(2)\)

Từ \((1);(2)\Rightarrow P\geq \frac{4(a+2b+1)}{a+2b+1}-2=2\) (đpcm)

Ta có:

\(\dfrac{x^2}{\sqrt{1-x^2}}=\dfrac{x^3}{x\sqrt{1-x^2}}\)

Áp dụng BĐT Cosi ta có:

\(x\sqrt{1-x^2}\le\dfrac{x^2+1-x^2}{2}=\dfrac{1}{2}\)

\(\Rightarrow\dfrac{x^3}{x\sqrt{1-x^2}}\ge2x^3\)

Cmtt:

\(\dfrac{y^3}{y\sqrt{1-y^2}}\ge2y^3\)

\(\dfrac{z^3}{z\sqrt{1-z^2}}\ge2z^3\)

\(\Rightarrow\dfrac{x^2}{\sqrt{1-x^2}}+\dfrac{y^2}{\sqrt{1-y^2}}+\dfrac{z^2}{\sqrt{1-z^2}}=\dfrac{x^3}{x\sqrt{1-x^2}}+\dfrac{y^3}{y\sqrt{1-y^2}}+\dfrac{z^3}{z\sqrt{1-z^2}}\ge2\left(x^3+y^3+z^3\right)=2\) (ĐPCM)

Đặt \(P=\left(\dfrac{x-y}{z}+\dfrac{y-z}{x}+\dfrac{z-x}{y}\right)\left(\dfrac{z}{x-y}+\dfrac{x}{y-z}+\dfrac{y}{z-x}\right)=9\)

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

\(\Leftrightarrow P=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\\ =1+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+1+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+1\\ =3+\dfrac{a+c}{b}+\dfrac{a+b}{c}+\dfrac{b+c}{a}\)

Ta có \(\dfrac{a+c}{b}=\dfrac{\dfrac{x-y}{z}+\dfrac{z-x}{y}}{\dfrac{y-z}{x}}=\dfrac{xy-y^2+z^2-xz}{yz}\cdot\dfrac{x}{y-z}\)

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

Mà \(x+y+z=0\Leftrightarrow x=-y-z\)

\(\Leftrightarrow\dfrac{a+c}{b}=\dfrac{x\left(x+x\right)}{yz}=\dfrac{2x^2}{yz}\)

Cmtt ta được \(\dfrac{a+b}{c}=\dfrac{2y^2}{xz};\dfrac{b+c}{a}=\dfrac{2z^2}{xy}\)

Cộng vế theo vế

\(\Leftrightarrow P=\dfrac{2x^2}{yz}+\dfrac{2y^2}{xz}+\dfrac{2z^2}{xy}+3=\dfrac{2x^3+2y^3+2z^3}{xyz}+3\\ \Leftrightarrow P=\dfrac{2\left(x^3+y^3+z^3\right)}{xyz}+3\)

Lại có \(x+y+z=0\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)=0\)

\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\\ \Leftrightarrow x^3+y^3+z^3=3xyz\)

Thế vào \(P\)

\(\Leftrightarrow P=\dfrac{2\cdot3xyz}{xyz}+3=6+3=9\)