Áp dụng BĐT phụ \(a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Leftrightarrow\left(a-b\right)^2\ge0\)

\(A\ge\dfrac{1}{2}\left(x+y+\dfrac{1}{x}+\dfrac{1}{y}\right)^2\ge\dfrac{1}{2}\left(x+y+\dfrac{4}{x+y}\right)^2=\dfrac{1}{2}\left(1+\dfrac{4}{1}\right)^2=\dfrac{25}{2}\)

Dấu "=" \(x=y=\dfrac{1}{2}\)

\(x+y+z=xyz\Leftrightarrow\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)

\(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\right)=2^2-2.1=2\) (đpcm)

\(3\left(x^2+y^2\right)-2\left(x^3+y^3\right)\)

\(=3\left[\left(x^2+2xy+y^2\right)-2xy\right]-2\left(x+y\right)\left(x^2-xy+y^2\right)\)

\(=3\left[\left(x+y\right)^2-2xy\right]-2\left[\left(x+y\right)^2-3xy\right]\)

\(=3\left(1-2xy\right)-2\left(1-3xy\right)\)

\(=6xy-6xy+3-2=1\)

Vậy với \(x+y=1\) thì \(3\left(x^2+y^2\right)-2\left(x^3+y^3\right)=1\)

kim ngân bn giải thích cho mk dòng thứ 3 :-3xy từ đâu có vậy??

\(x^2>=\dfrac{1}{4}\)

\(y^2>=\dfrac{1}{4}\)

Do đó: \(x^2+y^2>=\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{2}\)

\(x\ge\dfrac{1}{2};y\ge\dfrac{1}{2}\)=>\(xy\ge\dfrac{1}{4}\)=>\(2xy\ge\dfrac{1}{2}\).

\(x+y\ge\dfrac{1}{2}+\dfrac{1}{2}=1\)

=>\(\left(x+y\right)^2\ge1\)

=>\(x^2+2xy+y^2\ge1\)

=>\(x^2+y^2\ge1-2xy\ge1-\dfrac{1}{2}=\dfrac{1}{2}\)

Ta có \(x\ge\dfrac{1}{2},y\ge\dfrac{1}{2}\)

=>\(xy\ge\dfrac{1}{4}\)

Ta có :\(\left(x-y\right)^2\ge0\)

=>\(x^2-2xy+y^2\ge0\)

=>\(x^2-2\dfrac{1}{4}+y^2\ge0\)

=>\(x^2+y^2\ge\dfrac{1}{2}\)

** Bạn lưu ý lần sau viết đề bằng công thức toán!

Đề cần sửa thành $\leq \frac{4}{3}$

Lời giải:

Áp dụng BĐT AM-GM và Cauchy-Schwarz:

\(\frac{1}{2x^2+y^2+z^2}=\frac{1}{(x^2+z^2)+(x^2+y^2)}\leq \frac{1}{2xy+2xz}=\frac{1}{2}.\frac{1}{xy+xz}\leq \frac{1}{8}\left(\frac{1}{xy}+\frac{1}{xz}\right)\)

Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế suy ra:

\(\sum \frac{1}{2x^2+y^2+z^2}\leq \frac{1}{4}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)=\frac{x+y+z}{4xyz}\) $(1)$

Mặt khác:

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=4\Rightarrow 4xyz=xy+yz+xz$

$\Rightarrow 16x^2y^2z^2=(xy+yz+xz)^2\geq 3xyz(x+y+z)$ (theo BĐT AM-GM)

$\Rightarrow x+y+z\leq \frac{16}{3}xyz (2)$

Từ $(1);(2)\Rightarrow \sum \frac{1}{2x^2+y^2+z^2}\leq \frac{4}{3}$ 

Dấu "=" xảy ra khi $x=y=z=\frac{3}{4}$

\(\dfrac{1}{2x^2+y^2+z^2}=\dfrac{1}{x^2+y^2+x^2+z^2}\le\dfrac{1}{2xy+2xz}\le\dfrac{1}{8}\left(\dfrac{1}{xy}+\dfrac{1}{xz}\right)\)

Tương tự: \(\dfrac{1}{x^2+2y^2+z^2}\le\dfrac{1}{8}\left(\dfrac{1}{xy}+\dfrac{1}{yz}\right)\) ; \(\dfrac{1}{x^2+y^2+2z^2}\le\dfrac{1}{8}\left(\dfrac{1}{xz}+\dfrac{1}{yz}\right)\)

Cộng vế:

\(VT\le\dfrac{1}{4}\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\right)\le\dfrac{1}{4}.\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2=\dfrac{4}{3}\)

Đề bài sai