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

chứng minh là đề sai nhé :

\(2\sqrt{x}=1+\sqrt{y}\ge1\) \(\Rightarrow\sqrt{x}\ge\frac{1}{2}\Rightarrow x\ge\frac{1}{4}\)

\(x+y\ge\frac{1}{4}>\frac{1}{5}\)( ko có dấu bằng xảy ra )

mình nghĩ sửa \(2\sqrt{x}-\sqrt{y}=1\)thành \(2\sqrt{x}+\sqrt{y}=1\)

Khi đó: Áp dụng BĐT Bu-nhi-a-cốp-ski , ta có :

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

\(\Rightarrow x+y\ge\frac{1}{5}\) . Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\frac{2}{\sqrt{x}}=\frac{1}{\sqrt{y}}\\2\sqrt{x}+\sqrt{y}=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{4}{25}\\y=\frac{1}{25}\end{cases}}\)

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)

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

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 \(x=y=z\)

Áp dụng bđt \(\left(a^2+b^2\right)\left(c^2+d^2\right)\ge\left(ac+bd\right)^2\)

Dấu bằng xảy ra khi \(ad=bc\)

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

Dấu bằng xảy ra khi \(xy=\sqrt{1-x^2}.\sqrt{1-y^2}\Leftrightarrow x^2y^2=x^2y^2+1-\left(x^2+y^2\right)\)

\(\Leftrightarrow x^2+y^2=1\)

Ta có x + y + z = 1 nên z = 1 - x - y.

Bất đẳng thức cần chứng minh tương đương:

\(\dfrac{\sqrt{xy+z\left(x+y+z\right)}+\sqrt{2x^2+2y^2}}{1+\sqrt{xy}}\ge1\)

\(\Leftrightarrow\sqrt{\left(z+x\right)\left(z+y\right)}+\sqrt{2x^2+2y^2}\ge1+\sqrt{xy}\).

Áp dụng bất đẳng thức Cauchy - Schwarz:

\(\left(z+x\right)\left(z+y\right)\ge\left(\sqrt{z}.\sqrt{z}+\sqrt{x}.\sqrt{y}\right)^2=\left(z+\sqrt{xy}\right)^2\)

\(\Rightarrow\sqrt{\left(z+x\right)\left(z+y\right)}\ge z+\sqrt{xy}=\sqrt{xy}-x-y+1\); (1)

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

Cộng vế với vế của (1), (2) ta có đpcm.

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

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

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

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