làm r mà bạn ei

Ta có:

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

Đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}x=\sqrt{1-y^2}\\y=\sqrt{1-x^2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x^2=1-y^2\\y^2=1-x^2\end{matrix}\right.\)

\(\Rightarrow x^2+y^2=1\) (đpcm)

\(\left(x;y;z\right)=\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)\Rightarrow ab+bc+ca=2020\)

BĐT trở thành:

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+a+b+c+\sqrt{2020+a^2}+\sqrt{2020+b^2}+\sqrt{2020+c^2}\le\dfrac{2020.2021}{abc}\)

\(\Leftrightarrow\dfrac{ab+bc+ca}{abc}+a+b+c+\sqrt{2020+a^2}+\sqrt{2020+b^2}+\sqrt{2020+c^2}\le\dfrac{2020.2021}{abc}\)

\(\Leftrightarrow a+b+c+\sqrt{2020+a^2}+\sqrt{2020+b^2}+\sqrt{2020+c^2}\le\dfrac{2020^2}{abc}\)

Ta có: \(\sqrt{2020+a^2}=\sqrt{ab+bc+ca+a^2}=\sqrt{\left(a+b\right)\left(a+c\right)}\le\dfrac{1}{2}\left(2a+b+c\right)\)

Tương tự:...

\(\Rightarrow\sqrt{2020+a^2}+\sqrt{2020+b^2}+\sqrt{2020+c^2}\le2\left(a+b+c\right)\)

\(\Rightarrow a+b+c+\sqrt{2020+a^2}+\sqrt{2020+b^2}+\sqrt{2020+c^2}\le3\left(a+b+c\right)\)

Nên ta chỉ cần chứng minh:

\(3\left(a+b+c\right)\le\dfrac{2020^2}{abc}=\dfrac{\left(ab+bc+ca\right)^2}{abc}\)

\(\Leftrightarrow\left(ab+bc+ca\right)^2\ge3abc\left(a+b+c\right)\) (hiển nhiên đúng)

Dấu "=" xảy ra khi \(a=b=c\) hay \(x=y=z\)

Lời giải:

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

$\Leftrightarrow (x+\sqrt{x^2+1})(x-\sqrt{x^2+1})(y+\sqrt{y^2+1})=2(x-\sqrt{x^2+1})$

$\Leftrightarrow -(y+\sqrt{y^2+1})=2(x-\sqrt{x^2+1})$

$\Leftrightarrow 2x+\sqrt{y^2+1}=2\sqrt{x^2+1}-y$

$\Rightarrow (2x+\sqrt{y^2+1})^2=(2\sqrt{x^2+1}-y)^2$
$\Leftrightarrow 4x^2+y^2+1+4x\sqrt{y^2+1}=4(x^2+1)+y^2-4y\sqrt{x^2+1}$

$\Leftrightarrow 4(x\sqrt{y^2+1})+y\sqrt{x^2+1})=3$

$\Leftrightarrow 4Q=3$

$\Leftrightarrow Q=\frac{3}{4}$ 

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 \(x=y=\dfrac{1}{\sqrt{2}}\)

Với mọi x;y;z ta luôn có:

\(\left(x+y-1\right)^2+\left(z-\dfrac{1}{2}\right)^2\ge0\)

\(\Leftrightarrow x^2+y^2+2xy-2x-2y+1+z^2-z+\dfrac{1}{4}\ge0\)

\(\Leftrightarrow x^2+y^2+z^2+\dfrac{5}{4}+2xy-2x-2y-z\ge0\)

\(\Leftrightarrow2+2xy-2x-2y\ge z\)

\(\Leftrightarrow2\left(1-x\right)\left(1-y\right)\ge z\)

Dấu "=" xảy ra khi và chỉ khi \(x=y=z=\dfrac{1}{2}\)