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

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:

Áp dụng BĐT AM-GM:

$x^3+1=(x+1)(x^2-x+1)\leq \left(\frac{x+1+x^2-x+1}{2}\right)^2=\frac{(x^2+2)^2}{4}$

$\Rightarrow \sqrt{x^3+1}\leq \frac{x^2+2}{2}$

$\Rightarrow \frac{1}{\sqrt{x^3+1}}\geq \frac{2}{x^2+2}$. Tương tự với các phân thức khác và cộng theo vế:

$\sum \frac{1}{\sqrt{x^3+1}}\geq 2\sum \frac{1}{x^2+2}$

Áp dụng BĐT Cauchy-Schwarz:

$\sum \frac{1}{x^2+2}\geq \frac{9}{x^2+y^2+z^2+6}=\frac{9}{12+6}=\frac{1}{2}$

$\Rightarrow \sum \frac{1}{\sqrt{x^3+1}}\geq 2.\frac{1}{2}=1$
Ta có đpcm

Dấu "=" xảy ra khi $x=y=z=2$

Giả thiết thiếu rồi em, chỗ \(\dfrac{1}{x+1}+...\) thiếu đoạn sau nữa

gt <=>   \(x\sqrt{1-y^2}=1-y\sqrt{1-x^2}\)

<=>   \(x^2\left(1-y^2\right)=1+y^2\left(1-x^2\right)-2y\sqrt{1-x^2}\)

<=>   \(x^2-x^2y^2=1+y^2-x^2y^2-2y\sqrt{1-x^2}\)

<=>   \(2y\sqrt{1-x^2}=y^2-x^2+1\)

<=>   \(4y^2\left(1-x^2\right)=\left(y^2-x^2+1\right)^2\)

<=>   \(4y^2-4x^2y^2=x^4+y^4+1-2x^2y^2-2x^2+2y^2\)

<=>   \(x^4+y^4+2x^2y^2-2x^2-2y^2+1=0\)

<=>   \(\left(x^4+y^4+2x^2y^2\right)-2\left(x^2+y^2\right)+1=0\)

<=>   \(\left(x^2+y^2\right)^2-2\left(x^2+y^2\right)+1=0\)

<=>   \(\left(x^2+y^2-1\right)^2=0\)

<=>   \(x^2+y^2-1=0\)

<=>   \(x^2+y^2=1\)

VẬY TA CÓ ĐPCM.

Bài của Hermit thiếu điều kiện xác định + bài làm dài 

 \(-1\le x;y\le1\) theo bài ra ta có:

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

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

Dấu "=" xảy ra khi \(\hept{\begin{cases}\left|x\right|=\sqrt{1-y^2}\\\left|y\right|=\sqrt{1-x^2}\end{cases}\Leftrightarrow x^2=1-y^2\Leftrightarrow x^2+y^2=1\left(đpcm\right)}\)

Đặt \(\left(\dfrac{1}{\sqrt{x}};\dfrac{1}{\sqrt{y}};\dfrac{1}{\sqrt{z}}\right)=\left(a;b;c\right)\Rightarrow\dfrac{a^2}{a^2+1}+\dfrac{b^2}{b^2+1}+\dfrac{c^2}{c^2+1}=1\)

Ta cần chứng minh: \(ab+bc+ca\le\dfrac{3}{2}\)

Thật vậy, ta có:

\(1=\dfrac{a^2}{a^2+1}+\dfrac{b^2}{b^2+1}+\dfrac{c^2}{c^2+1}\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3}\)

\(\Rightarrow a^2+b^2+c^2+3\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)

\(\Rightarrow ab+bc+ca\le\dfrac{3}{2}\) (đpcm)

\(\dfrac{\sqrt{1+x^3+y^3}}{xy}>=\sqrt{\dfrac{3}{xy}}\)

\(\dfrac{\sqrt{1+y^3+z^3}}{yz}>=\sqrt{\dfrac{3}{yz}}\)

\(\dfrac{\sqrt{1+z^3+x^3}}{xz}>=\sqrt{\dfrac{3}{xz}}\)

=>\(VT>=\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)=3\sqrt{3}\)