Áp dụng BĐT AM-GM: \(\left(x^2-y^2\right)\left(1-x^2y^2\right)\le\frac{1}{4}\left(x^2-y^2+1-x^2y^2\right)^2=\frac{1}{4}\left(1-y^2\right)^2\left(1+x^2\right)^2\)

\(P\le\frac{1}{4}\frac{\left(1-y^2\right)^2}{\left(1+y^2\right)^2}\)

mà theo BĐT AM-GM:\(\left(1-y\right)\left(1+y\right)\le\frac{1}{4}\left(1-y+1+y\right)^2=1\)

\(\Rightarrow P\le\frac{1}{4}.\frac{1}{\left(1+y^2\right)^2}\le\frac{1}{4}.\frac{1}{1}=\frac{1}{4}\)

Dấu = xảy ra khi x=1;y=0 wait : có gì đó sai sai. số thực

Áp dụng bất đẳng thức Cauchy, ta có: \(\sqrt{x\left(2x+y\right)}=\frac{1}{\sqrt{3}}.\sqrt{3x\left(2x+y\right)}\le\frac{5x+y}{2\sqrt{3}}\)

Tương tự: \(\sqrt{y\left(2y+x\right)}\le\frac{5y+x}{2\sqrt{3}}\)

\(\Rightarrow\sqrt{x\left(2x+y\right)}+\sqrt{y\left(2y+x\right)}\le\frac{6\left(x+y\right)}{2\sqrt{3}}=\frac{3\left(x+y\right)}{\sqrt{3}}\)\(\Rightarrow P=\frac{x+y}{\sqrt{x\left(2x+y\right)}+\sqrt{y\left(2y+x\right)}}\ge\frac{x+y}{\frac{3}{\sqrt{3}}\left(x+y\right)}=\frac{1}{\sqrt{3}}\)

Đẳng thức xảy ra khi x = y

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

\(\ge\frac{2}{\sqrt{xy}}\sqrt{1+x^2y^2}=2\sqrt{\frac{1}{xy}+xy}=2\sqrt{\frac{1}{16xy}+xy+\frac{15}{16xy}}\)

\(\ge2\sqrt{2\sqrt{\frac{1}{16xy}\cdot xy}+\frac{15}{4\left(x+y\right)^2}}=2\sqrt{\frac{1}{2}+\frac{15}{4}}=\sqrt{17}\)

Dấu "=" xảy ra tai x=y=1/2

\(C=\left|2x+1\right|+\left|-2y-1\right|\ge\left|2x+1-2y-1\right|=2\left|x-y\right|=4\)

\(C_{min}=4\) 

\(\Leftrightarrow2y^3-6y^2+7y-3=-2x\sqrt{1-x}+2\sqrt{1-x}+\sqrt{1-x}\)

\(\Leftrightarrow2\left(y^3-3y^2+3y+1\right)+y-1=2\left(1-x\right)\sqrt{1-x}+\sqrt{1-x}\)

\(\Leftrightarrow2\left(y-1\right)^3+y-1=2\left(\sqrt{1-x}\right)^3+\sqrt{1-x}\) (1)

Xét hàm \(f\left(t\right)=2t^3+t\)

\(f'\left(t\right)=6t^2+1>0\Rightarrow f\left(t\right)\) đồng biến

Nên (1) tương đương: \(y-1=\sqrt{1-x}\Rightarrow y=1+\sqrt{1-x}\)

\(\Rightarrow P=x+2\sqrt{1-x}+2=-\left(1-x-2\sqrt{1-x}+1\right)+4=-\left(\sqrt{1-x}-1\right)^2+4\le4\)

⇒ P = x + 2 √ 1 − x + 2

= − ( 1 − x − 2 √ 1 − x + 1 ) + 4

= − ( √ 1 − x − 1 ) 2 + 4 ≤ 4

Cho xin một like đi các dân chơi à.

Bài 2: 

Tìm GTLN: \(x^2+xy+y^2=3\Leftrightarrow xy=\left(x+y\right)^2-3\Rightarrow xy\ge-3\Rightarrow-7xy\le21\)

\(P=2\left(x^2+xy+y^2\right)-7xy\le2.3+21=27\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x+y=0\\xy=-3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\sqrt{3},y=-\sqrt{3}\\x=-\sqrt{3},y=\sqrt{3}\end{cases}}\)

Tìm GTNN: 

 Chứng minh \(xy\le\frac{1}{2}\left(x^2+y^2\right)\Rightarrow\frac{3}{2}xy\le\frac{1}{2}\left(x^2+y^2+xy\right)\)

\(\Rightarrow\frac{3}{2}xy\le\frac{3}{2}\Rightarrow xy\le1\Rightarrow-7xy\ge-7\)

\(P=2\left(x^2+xy+y^2\right)-7xy\ge2.3-7=-1\)

Chúc bạn học tốt.

Làm bài 1 ha :) 

Áp dụng BĐT Cô si ta có:

\(\left(1-x^3\right)+\left(1-y^3\right)+\left(1-z^3\right)\ge3\sqrt[3]{\left(1-x^3\right)\left(1-y^3\right)\left(1-z^3\right)}\)

\(\Leftrightarrow\frac{3-\left(x^3+y^3+z^3\right)}{3}\ge\sqrt[3]{\left(1-x^3\right)\left(1-y^3\right)\left(1-z^3\right)}\)

Mặt khác:\(\frac{3-\left(x^3+y^3+z^3\right)}{3}\le\frac{3-3xyz}{3}=1-xyz\)

Khi đó:

\(\left(1-xyz\right)^3\ge\left(1-x^3\right)\left(1-y^3\right)\left(1-z^3\right)\)

Giống Holder ghê vậy ta :D

Max=3,222222