\(\left(2\sqrt{1+a}\right)^2=4\left(1+a\right)=\left(\sqrt{1+x}+\sqrt{1+y}\right)^2\le2\left(x+y+2\right)\)

\(\Leftrightarrow\)\(x+y\ge2a\)

Áp dụng bđt Bunyakovsky: \(\left(\sqrt{1+x}+\sqrt{1+y}\right)^2\le2\left(x+y+2\right)\)

\(\Rightarrow4\left(a+1\right)\le2\left(x+y+2\right)\Leftrightarrow4a\le2\left(x+y\right)\Leftrightarrow x+y\ge2a\)

Áp dụng bất đẳng thức Schwartz , ta có : 

\(\left(1.\sqrt{1+x}+1.\sqrt{1+y}\right)^2\le\left(1^2+1^2\right)\left(1+x+1+y\right)\)

\(\Leftrightarrow4\left(1+a\right)\le2.\left(x+y+2\right)\)

\(\Leftrightarrow x+y+2\ge2a+2\)

\(\Rightarrow x+y\ge2a\left(ĐPCM\right)\)

Cho kq luôn :X=1

Áp dụng bất đẳng thức Bunhiacopxki ta có:

\(\left(x\cdot1+y\cdot1\right)^2\le\left(1^2+1^2\right)\left(x^2+y^2\right)=2\Rightarrow x+y\le\sqrt{2}\)

Áp dụng bất đẳng thức Bunhiacopxki ta có:

\(\left(x\sqrt{1+y}+y\sqrt{1+x}\right)^2\le\left(x^2+y^2\right)\left(1+y+1+x\right)=x+y+2=2+\sqrt{2}\)

\(\Rightarrow x\sqrt{y+1}+y\sqrt{x+1}\ge\sqrt{2+\sqrt{2}}\)

Dấu = xảy ra khi \(x=y=\dfrac{1}{\sqrt{2}}\)

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

\(\Rightarrow2\sqrt{1+x^2}.y'=\dfrac{2\sqrt{1+x^2}\left(1+\dfrac{x}{\sqrt{1+x^2}}\right)}{2\sqrt{x+\sqrt{1+x^2}}}\)

\(=\dfrac{\sqrt{1+x^2}+x}{\sqrt{x+\sqrt{1+x^2}}}=\sqrt{x+\sqrt{1+x^2}}=y\) (đpcm)

Xét \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=2\)

\(\Leftrightarrow1=\left(1-\dfrac{1}{x}\right)+\left(1-\dfrac{1}{y}\right)+\left(1-\dfrac{1}{z}\right)\)

\(\Leftrightarrow1=\dfrac{x-1}{x}+\dfrac{y-1}{y}+\dfrac{z-1}{z}\)

Áp dụng bđt Bunhiacopxki có:

\(x+y+z=\left(x+y+z\right)\left(\dfrac{x-1}{x}+\dfrac{y-1}{y}+\dfrac{z-1}{1}\right)\ge\left(\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}\right)^2\)\(\Leftrightarrow\sqrt{x+y+z}\ge\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}\)

Dấu "=" xảy ra khi x=y=z=1,5Tự đăng câu hỏi xong tự trả lời  (T-T)    

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:

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

Lời giải:

a) ĐK: $x\geq 0; y\geq 0; x\neq y$

\(A=\left[\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}{\sqrt{x}-\sqrt{y}}-\frac{(\sqrt{x}-\sqrt{y})(x+\sqrt{xy}+y)}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}\right]:\frac{x-\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)

\(=\left(\sqrt{x}+\sqrt{y}-\frac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right).\frac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)

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

b) \(1-A=\frac{(\sqrt{x}-\sqrt{y})^2}{x-\sqrt{xy}+y}>0\) với mọi $x\neq y; x,y\geq 0$

$\Rightarrow A< 1$

ĐỊT MẸ