\(x+2y=\sqrt{\left(\frac{1}{\sqrt{2}}.\sqrt{2}x+\frac{2}{\sqrt{3}}.\sqrt{3}y\right)^2}\le\sqrt{\left(\frac{1}{2}+\frac{4}{3}\right)\left(2x^2+3y^2\right)}=\sqrt{\frac{22}{3}}\)

Áp dụng BĐT Bunhicopxki:

\(\left(\sqrt{\frac{1}{2}}^2+\sqrt{\frac{4}{3}}^2\right)\left(\left(\sqrt{2}x\right)^2+\left(\sqrt{3}y\right)^2\right)\ge\left(x+2y\right)^2\)

\(\Leftrightarrow\frac{11}{6}\left(2x^2+3y^2\right)\ge\left(x+2y\right)^2\)

\(\Leftrightarrow\frac{44}{6}=\frac{22}{3}\ge\left(x+2y\right)^2\)(1)

Do x, y > 0 nên x + 2y > 0 do đó từ (1) suy ra \(x+2y\le\sqrt{\frac{22}{3}}\)(đpcm)

bạn xem câu hỏi số 905663 nhé

Đề kì vậy bạn. Sao vế trái không có \(y\) vậy?

Liên tục áp dụng bất đẳng thức \(\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\) và ta có:

\(\frac{1}{3x+3y+2x}=\frac{1}{2\left(x+y\right)+\left(x+y+2z\right)}\le\frac{1}{4}\left(\frac{1}{2\left(x+y\right)}+\frac{1}{\left(x+z\right)+\left(y+z\right)}\right)\le\frac{1}{8\left(x+y\right)}+\frac{1}{16}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\)

Chứng minh tương tự tạ có:

\(\frac{1}{3x+2y+3z}\le\frac{1}{8\left(z+x\right)}+\frac{1}{16}\left(\frac{1}{x+y}+\frac{1}{y+z}\right)\)

\(\frac{1}{2x+3y+3z}\le\frac{1}{8\left(y+z\right)}+\frac{1}{16}\left(\frac{1}{z+x}+\frac{1}{x+y}\right)\)

Suy ra \(VT\le\frac{1}{8}\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)+\frac{1}{8}\left(\frac{1}{x+y}+\frac{1}{x+z}+\frac{1}{z+x}\right)=\frac{3}{2}\)

Dấu "=" xảy ra <=> \(x=y=z=\frac{1}{4}\)

@Lightning Farron

Với $x+y \geqslant 0$, ta có:

$2x^2+2y^2 \geqslant (x+y)^2 \Rightarrow \sqrt{2x^2+2y^2} \geqslant x+y$

\(x^2+xy+y^2=(x+y)^2-xy \geqslant (x+y)^2-\dfrac{(x+y)^2}{4} \Rightarrow \sqrt {\dfrac{{4\left( {{x^2} + xy + {y^2}} \right)}}{3}} \ge x + y\)

$\sqrt{2x^2+2y^2}+\sqrt {\dfrac{{4\left( {{x^2} + xy + {y^2}} \right)}}{3}} \geqslant 2(x+y) \Rightarrow PT(2) \Leftrightarrow x = y$

Vậy hệ phương trình có 2 nghiệm $(x;y)$ là $(0;0); (1;1)$