ADBDT Cauchy:

2(x^2+y^2)>=(x+y)^2

Dau = khi x=y

Lời giải;

Vế 1:

Áp dụng BĐT AM-GM:

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

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

$y^3+\frac{y}{2}\geq \sqrt{2}y^2$

$\Rightarrow x^3+y^3+\frac{x+y}{2}\geq \sqrt{2}(x^2+y^2)=\sqrt{2}$

$\Rightarrow x^3+y^3\geq \sqrt{2}-\frac{x+y}{2}\geq \sqrt{2}-\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$

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Vế 2:

$x^2+y^2=1$

$\Rightarrow x^2=1-y^2\leq 1\Rightarrow -1\leq x\leq 1$

$y^2=1-x^2\leq 1\Rightarrow -1\leq y\leq 1$

$\Rightarrow x^3\leq x^2; y^3\leq y^2$

$\Rightarrow x^3+y^3\leq x^2+y^2$ hay $x^3+y^3\leq 1$

Lời giải:
Áp dụng BĐT AM-GM:
$2A=2x^2y^2(x^2+y^2)=xy.[2xy(x^2+y^2)]\leq \left(\frac{x+y}{2}\right)^2.\left(\frac{2xy+x^2+y^2}{2}\right)^2$

$\Leftrightarrow 2A\leq \frac{(x+y)^6}{16}=\frac{1}{16}$

$\Rightarrow A\leq \frac{1}{32}$
Vậy $A_{\max}=\frac{1}{32}$. Giá trị này đạt được khi $x=y=\frac{1}{2}$

Áp dụng bdt cosi-schwar cho 3 số (\(\left(am+bn+cp\right)^2\le\left(a^2+b^2+c^2\right)\)\(\left(m^2+n^2+p^2\right)\)

với a=x,b=y\(\sqrt{2}\);c=z\(\sqrt{5}\);  m=\(\sqrt{11-2y^2},n=\sqrt{3-5z^2}\),\(p=\sqrt{2-x^2}\)

8²\(\le\left(x^2+2y^2+5z^2\right)\left(11-2y^2+3-5z^2+1-x^2\right)\)  <=>64\(\le P\left(16-P\right)\)

<=>P²-16P+64\(\le0< =>\left(P-8\right)^2\le0\)  <=>P=8

Đề bài sai, sửa đề: \(2\le\sqrt{x^2+y^2}+\sqrt{xy}\le\sqrt{6}\)

Đặt \(P=\sqrt{x^2+y^2}+\sqrt{xy}>0\)

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

\(\Rightarrow P^2\ge x^2+y^2+\left(2\sqrt{2}+1\right)xy\ge x^2+y^2+2xy=4\)

\(\Rightarrow P\ge2\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(2;0\right);\left(0;2\right)\)

Lại có: \(P^2=x^2+y^2+xy+2\sqrt{\left(x^2+y^2\right)xy}=x^2+y^2+xy+\sqrt{4xy.\left(x^2+y^2\right)}\)

\(\Rightarrow P^2\le x^2+y^2+xy+\dfrac{1}{2}\left(4xy+x^2+y^2\right)=\dfrac{3}{2}\left(x+y\right)^2=6\)

\(\Rightarrow P\le\sqrt{6}\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(\dfrac{3-\sqrt{3}}{3};\dfrac{3+\sqrt{3}}{3}\right)\)