Đặt \(\dfrac{x}{z}=a;\dfrac{y}{z}=b\).

Theo gt ta có \(a+b\le1\).

BĐT cần chứng minh tương đương:

\(a^2+b^2+\frac{a^2}{b^2}+\frac{b^2}{a^2}+\frac{1}{a^2}+\frac{1}{b^2}\ge \frac{21}{2}\).

Theo bđt AM - GM: \(\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}\ge2;a^2+\dfrac{1}{16}a^2\ge\dfrac{1}{2};b^2+\dfrac{1}{16}b^2\ge\dfrac{1}{2};\dfrac{15}{16}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)\ge\dfrac{15}{32}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^2\ge\dfrac{15}{32}.\left(\dfrac{4}{a+b}\right)^2\ge\dfrac{15}{2}\).

Cộng vế với vế của các bđt trên lại ta có đpcm.

Thay $x=\sqrt{\frac{1}{2,5}}; y=z=\sqrt{\frac{1}{0,25}}$ ta thấy đề sai bạn nhé!

\(VT=\dfrac{1}{\left(x-y\right)^2}+\dfrac{x^2+y^2}{x^2y^2}=\dfrac{1}{\left(x-y\right)^2}+\dfrac{\left(x-y\right)^2+2xy}{x^2y^2}\)

\(VT=\dfrac{1}{\left(x-y\right)^2}+\dfrac{\left(x-y\right)^2}{x^2y^2}+\dfrac{2}{xy}\ge2\sqrt{\dfrac{\left(x-y\right)^2}{\left(x-y\right)^2x^2y^2}}+\dfrac{2}{xy}=\dfrac{2}{\left|xy\right|}+\dfrac{2}{xy}\ge\dfrac{2}{xy}+\dfrac{2}{xy}=\dfrac{4}{xy}\)

\(\left(1+x\right)^2=\left(1.1+\sqrt{xy}.\sqrt{\dfrac{x}{y}}\right)^2\le\left(1+xy\right)\left(1+\dfrac{x}{y}\right)=\dfrac{\left(1+xy\right)\left(x+y\right)}{y}\)

\(\Rightarrow\dfrac{1}{\left(1+x\right)^2}\ge\dfrac{y}{\left(1+xy\right)\left(x+y\right)}\)

Tương tự ta có: \(\dfrac{1}{\left(1+y\right)^2}\ge\dfrac{x}{\left(1+xy\right)\left(x+y\right)}\)

Cộng vế với vế: 

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

Dấu "=" xảy ra khi \(x=y=1\)

Lời giải:
Áp dụng BĐT Bunhiacopxky:

$\text{VT}(1^2+1^2+1^2)\geq (1+\frac{x}{y+z}+1+\frac{y}{x+z}+1+\frac{z}{x+y})^2$

$\Leftrightarrow 3\text{VT}\geq (3+\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y})^2$

$ = \left[3+\frac{x^2}{xy+xz}+\frac{y^2}{yz+yx}+\frac{z^2}{zy+zx}\right]^2$

$\geq \left[3+\frac{(x+y+z)^2}{2(xy+yz+xz)}\right]^2$

$\geq \left[3+\frac{3(xy+yz+xz)}{2(xy+yz+xz)}\right]^2=\frac{81}{4}$

$\Rightarrow \text{VT}\geq \frac{27}{4}$

Dấu "=" xảy ra khi $x=y=z>0$

Áp dụng BĐT Bunhiacopxky:

VT(12+12+12)≥(1+��+�+1+��+�+1+��+�)2VT(12+12+12)≥(1+y+zx​+1+x+zy​+1+x+yz​)2

⇔3VT≥(3+��+�+��+�+��+�)2⇔3VT≥(3+y+zx​+x+zy​+x+yz​)2

=[3+�2��+��+�2��+��+�2��+��]2=[3+xy+xzx2​+yz+yxy2​+zy+zxz2​]2

≥[3+(�+�+�)22(��+��+��)]2≥[3+2(xy+yz+xz)(x+y+z)2​]2

≥[3+3(��+��+��)2(��+��+��)]2=814≥[3+2(xy+yz+xz)3(xy+yz+xz

Gợi ý: \(\dfrac{a^4+b^4}{2}\ge\left(\dfrac{a+b}{2}\right)^4\)

Ta có \(a^4+b^4\ge\dfrac{\left(a^2+b^2\right)^2}{2}\ge\dfrac{\left(\dfrac{\left(a+b\right)^2}{2}\right)^2}{2}=\dfrac{\left(a+b\right)^4}{8}\). Áp dụng cho biểu thức A, suy ra \(A\ge\dfrac{\left(x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}+2\right)^4}{8}\). Ta tìm GTNN của \(P=x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}+2\). Ta có 

\(P=x^2+\dfrac{1}{16x^2}+y^2+\dfrac{1}{16y^2}+\dfrac{15}{16}\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+2\)

\(P\ge2\sqrt{x^2.\dfrac{1}{16x^2}}+2\sqrt{y^2.\dfrac{1}{16y^2}}+\dfrac{15}{16}\left(\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2}{2}\right)+2\)

    \(=\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{15}{16}.\left(\dfrac{4^2}{2}\right)+2\) \(=\dfrac{21}{2}\). Do đó \(P\ge\dfrac{21}{2}\) \(\Leftrightarrow A\ge\dfrac{\left(\dfrac{17}{2}+2\right)^4}{8}\). Vậy GTNN của A là \(\dfrac{\left(\dfrac{17}{2}+2\right)^4}{8}\), ĐTXR \(\Leftrightarrow x=y=\dfrac{1}{2}\)